Excuse my little bit of fun.

I have started reading the english translation of kiselev's book and i have an inquiry as to the meaning of one of the very first statements on the introduction in page 1.

the statement concerning the definition of a surface and i quote:

"A geometric solid is separated from the surrounding space by a surface".

I might be completely off here with my reasoning or my trying to understand what is clearly meant by that statement since English is not my first language.

If i may present what i mean by me finding the meaning of the statement a bit ambiguous.

Prior to this statement a definition of what constitutes a geometric solid is provided such that:

" the part of space occupied by a physical object is called a geometric solid"

Which leds me to think that a geometric solid occupying physical space manifests itself in 3 dimensions namely length, width or thickness (x,y,z) otherwise it could not be occupying physical space and as such cannot be declared to be a geometric solid. Following from that anything that does not manifest itself in (x,y,z) is not a geometric solid.

I am hoping or assuming that my reasoning so far is correct.

Now, when we do have a geometric solid folowing from kiselovs statement what separates the surrounding space from the solid is a surface. Now geometric solids other than the shpere have more than one sides or surface? Would surface refer to the sum of those sides or a surface would refer to one of the sides of the geometric solid at any one time?

Thank you for your time.

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Forum thread: Angles ]]>

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[In fact there is some ambiguity in this problem, for - which of the angles between two tangents is 60 degrees?]

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Forum thread: Inscribed And Some Other Angles ]]>

Can I get some clarification please? There was a remark appended to the problem which clarified the problem a bit, but I'm still not completely certain about what is required to be found, nor what is given.

Thanks

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Forum thread: Inscribed And Some Other Angles ]]>

Do I understand this correctly? Let the given segment be 'a'. If I choose a point P in the plane, and draw two tangents to some circle, and mark the tangency points A and B, then if PA = PB = 'a', the point P satisfies the condition for membership of the locus of points?

If that's the case, then the problem is quite easy once one learns the solution for exercise 249.

Writing this helped me understand the nature of the problem.

I've completed all of the problems in this chapter except for the very first. Once I have some time (probably 2-3 weeks from now), I'll dedicate some time to posting solutions to exercises in chapters 2.1 and 2.2. So far, I've had a very great time learning elementary geometry with this book. If a solution set to all (or nearly all) problems existed, then I think that this would be more readily adopted by more educators, and therefore more students would receive a rich elementary geometry experience.

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2. I see it. The solution for AM < AP using the triangle inequality involved drawing 1 less auxiliary line, and was seen at a glance.

3. I incorrectly self-imposed this condition. As Polya said, the first step to problem-solving is to understand the problem. The solution for A outside the disk can use a similar angle-based argument—it's slightly different, so I'll take some time to post it for other readers later.

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Forum thread: Circles And Chords ]]>

2. Your argument to show that AM > AQ works, and it is also possible to show that AM<AP by the same argument, based on angles. [Why do you use different arguments for proving similar statements?]

Yet, what if A is outside the circle? (Your argument based on angles should also work, but might appear different due to a new disposition of the points involved).

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Forum thread: Circles And Chords ]]>

Let a circle with centre O be given, and an arbitrary point A in its interior region also be given. Draw a secant through O and A, and mark its intersections with the circle as Q and P. Choose an arbitrary point M along the circle not equal to P or Q, and draw a line from it through A up to its intersection with the circle at N.

1. (part 48) NA < OA + ON, and ON = PO as radii of the same circle. Therefore, NA < OA + PO, and thus, because PO + OA are summands of the segment PA, and N is arbitrary, P is the farthest point from A.

2. Draw a line from M to O, then OM = OQ as radii of the same circle, so the triangle OQM is isosceles, and thus (part 35) the angles at the base are congruent (i.e. OQM = OMQ). But then in the triangle AQM, the angle AMQ < (AQM = OQM) because AMQ is a part of angle OMQ. Therefore, (part 45) AM > AQ as a side opposite to a greater angle in a triangle. Because M was chosen to be arbitrary, Q is the closest point on the circle to A.

It doesn't sound like my argument for point 2 was what you had in mind. Seemed a bit easy for an asterisked question. I'll see if I can approach the problem from another perspective. If it's possible to boost my karma, then I would appreciate it if you would do that so that I can post images here. Thanks.

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I believe, if you make an accurate argument (based on this hint) for the farthest point, you'll see that a similar argument works for the closest point as well. Good luck!

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Solution:

Let angle BAC be given, and a point Q in its interior also be given. Mark the point D symmetric to Q about AB, and mark the point E symmetric to Q about AC. Draw a line from D to E, and mark the points where it intersects AB and AC as M and N, respectively. Draw lines from Q to both M and N, and from M to N. The triangle QMN is the required construction, ie. that triangle which satisfies the given conditions of the problem.

Proof: Take an arbitrary on each side of angle BAC except both M and N concurrently, and mark the one on AB as J, and the one on AC as K. Join J to D and K to E by lines. Notice that QJK = DJ + JK + KE (by the properties of axial symmetry), and this broken line is smallest when it is straight (part 49), ie. DM + MN + NE = QMN.

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Forum thread: Methods Of Construction And Symmetries ]]>

(1) If a is less than the distance from PQ to MM (as defined by part 85), then there is no solution because any perpendicular dropped from PQ to MM (or vice-versa) would be smaller than any slant (part 51), and a will only form a part of the perpendicular.

(2) If a is not less than the distance from PQ to MM, then it can be constructed through O, with endpoints on PQ and MM, respectively, using the following process:

Choose a point A along MN, and setting the compass to a step equal to a, place the pin leg at A. Using the compass' pencil leg, describe an arc of radius a, and mark its intersection with PQ as B.

Draw a line from A to B. If O lies along A and B, then we are done, so suppose it does not. Then, through the point O, construct a line parallel to AB, and mark its intersections with PQ and MN as C and D, respectively.

ABCD is a parallelogram (because by hypothesis, PQ || MN, and this implies CB || DA as segments of respectively parallel lines, and CD || AB by construction), so by part 85, CD = AB = *a* as opposite sides of a parallelogram.

But O belongs to CD by construction, and so we have constructed the required segment.

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Forum thread: General ]]>

It's taught using the first book published by the AMS. The course (Math 223: Linear Algebra) is taught at UBC Vancouver, and it's taught by Jozsef Solymosi.

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here and corresponds to a junior-level, upper-division course in Linear Algebra. It is formally independent of the first one, and at the beginning contains a couple of sections about vectors in elementary geometry essentially copied from my adaptation of Kiselev's "Stereometry".

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Forum thread: General ]]>

I'm considering registering for a course which teaches using your book, but my only meaningful contact with geometry will have been through the (expected by that time) completion of Kiselev's Geometry Book I. Planimetry. Should I postpone the course until I've completed Book II. Stereometry? Because of the existence of your translations of Kiselev's series about geometry, I suspect that a course based on your writings might come with some high expectations (concerning Euclidian geometry); without meeting those expectations, I suspect that a student cannot master the material.

Thank you.

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Forum thread: General ]]>

This can be interpreted as saying that the diagonals bisect each other, which (by part 87 (2)) means that the figure is a parallelogram.

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Then, we can assert that the altitude BF would be congruent to any altitude drawn to a lateral side of DBE (the proof of this is easy since the triangle is equilateral. I don't know if we've proven this prior to this exercise, so I don't have any theorem to reference); this justifies the application of Exercise 187 to assert congruence of BF and MX + MY.

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Alternative approach: Drop perpendiculars from C to EF, and F to AD; mark their feet as X and Y, respectively. Angles D and CFJ are congruent as corresponding angles, and CF = FD by definition of the midline. Therefore, triangles CFX and FDY are congruent (either by SAA or by part 55 (1)), so CX = FY. Drop a perpendicular from G to EF and H to EF, and mark their feet as Q and P respectively. Then GQ || CX (by part 71, as two perpendiculars dropped to the same line), GQ || HP for the same reason, and HP || FY too. Then, by part 85, HP = FY = CX = GQ. Also, angles GJQ and HJP are congruent (as vertical angles), and angles JHP and JGQ are congruent (as alternate angles). Therefore, by SAA, triangles JHP and GJQ are congruent, which implies the congruence of GJ and JH.

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a)

Let the given sides be $AB$, $BC$, and the median be $BD$. $D$ is the midpoint along the unknown third side. In the original triangle there is a line segment $ED$ from the midpoint, $E$, of side $BC$ to the midpoint of the third side; $ED = \frac{1}{2}AB$. $\S 95$) triangle midline theorem.

Bisect given sides $AB$ and $BC$, midpoints $F$ and $E$ respectively. Using the method in $\S 62$ construct triangle $D'E'B'$ using sides $AF$, $BD$, and $BE$. Using a straightedge and compass, extend $BE$ in the direction of $E$, set the compass step congruent to $BC$, place the pin at $B'$, sweep in the direction of $E'$ until it intersects $B'E'$ at point $C'$.

Construct segment $D'C'$ and extend sufficiently beyond $D'$ in the direction of $D'$. Set the compass step congruent to $AB$, place the pin at $B'$, and sweep until it intersects $D'C'$ in the direction away from $D'C'$; intersection point $A'$. Construct segment $A'B'$. $\triangle A'B'C'$ is the required triangle.

b)

Let the given base be $AC$, media $ZE$, and altitude $BD$. Construct base $A'C' = AC$. Use the method in $\S 67$ to bisect $A'C'$ and $BD$, midpoints $M'$ and $N$ respectively. Use the method in $\S 65$ to erect a perpendicular to $A'C'$ at $A'$, perpendicular $A'Y$. Set the compass step congruent to $BN$, place the pin at $A'$, sweep to intersect $A'Y$; intersection point $N'$.

Erect a perpendicular to $A'N'$ at $N'$ and in the plane in the direction of $C'$, perpendicular $N'X$. Set the compass step congruent to median $ZE$, place the pin at $A'$, and sweep to intersect $N'X$; intersection point $E'$. Construct line $C'E'$. $E'M'$ is congruent to half the triangle lateral side originating at $A'$, $\S 95$) triangle midline theorem.

Construct another line $PQ$ for the purpose of doubling $E'M'$. Set the compass step congruent to $E'M'$, place the pin at $P$, sweep in the direction of $Q$, intersection is $R$. Place compass pin at $R$ and sweep in the direction of $Q$, intersection point $S$. $PS = 2(E'M') =$ the triangle lateral side origin $A$. Set the compass congruent to $PS$, place the pin at $A'$, and sweep in the direction of $E'C'$; intersection point is $B'$. $\triangle A'B'C'$.

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Forum thread: Methods Of Construction And Symmetries ]]>

X

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Forum thread: Methods Of Construction And Symmetries ]]>

Let the given diagonals be $AB$ and $CD$, and the altitude be $EF$. Use the method in $\S 67$ to bisect $AB$ and $CD$, midpoints $J$ and $K$ respectively. Use a straightedge to construct a line $XY$. At an arbitrary point $E'$ along $XY$, use the method in $\S 65$ to erect a perpendicular to $XY$.

Set the compass step congruent to $EF$, place the pin at $E'$, and mark segment $E'C'$ along this perpendicular. Use the method in $\S 65$ to construct a perpendicular to $E'C'$ at $C'$, perpendicular $C'Q$. Set the compass step congruent to $CD$, place the pin at $C'$, and sweep until the intersection with $XY$, point $D'$. Construct segment $C'D'$. $\S 87(1)$ parallelogram diagonals bisect each other.

Set the compass step congruent to $CK$, place the pin at $C'$, and mark segment $C'K'$ along $C'D'$ in the direction of $D'$; $K'$ is the midpoint. Use the method in $\S 66$ to drop a perpendicular from $XY$ to $K'$, foot is $P$. Set the compass congruent to $AJ$, place the pin at $K'$, and sweep to mark intersections $A'$ with $XY$(in the same side of the plane about $PK'$ as $C'$) and $B'$ with $C'Q$(in the same side of the plane about $K'P$ as $D'$).

Construct segments $A'B'$, $A'C'$, $D'B'$. $CD = C'D' =$ given diagonal #2, by construction. $A'D' \parallel C'B'$ by construction. $\angle B'C'D' = \angle A'D'C'$ and $\angle A'C'D' = \angle B'D'C'$; $\S 77(2)$ parallels transversed alternate angles are congruent. $\triangle A'C'D' = \triangle B'C'D'$, $\S 40(2) ASA$. $\angle C'A'D' = \angle C'B'D'$, $C'B' = A'D'$.

$A'C'B'D'$ is a parallelogram, $\S 86(2)$ quadrilateral with two opposite sides parallel and congruent. $A'B' = AB =$ given diagonal #1. $C'E' = EF =$ given altitude and by construction it is perpendicular to the bases, therefore the altitude; $\S 85$) parallelogram altitude is line segment connecting the bases and perpendicular to them.

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Forum thread: Methods Of Construction And Symmetries ]]>

(a)

Let the given sum of the diagonal and side be $AC$. Initially assume that it is known the point of $AC$ that is the side, $AB$, and the part that is the diagonal, $BC$. Use the method in $\S 63$ to construct an angle congruent to the angle $\frac{1}{2}d$ with vertex at point $C$ and on side along $CA$; the other side is $CY$. Use a straightedge and compass to make a segment $CG$ along $CY$ congruent to $AB$. Continuing in the direction of $Y$, mark a segment $GH$ congruent to $BC$. Use the method in $\S 63$ to construct an angle congruent to the angle $\frac{1}{2}d$ with vertex $A$, a side along $AC$, and the other side in the part of the plane containing $Y$; the latter side is $AF$ and the intersection with $CH$ is point $J$.

Use the method in $\S 63$ to construct an angle at $B$ congruent to $\frac{1}{2}d$ with one side along $BC$ and the other, $BG$, in the direction of $CY$; the intersection with $CY$ is $G$. $\angle CBG = \frac{1}{2}d = \angle ACH$. $\triangle BGC$ is isoceles, $\S 45(1)$ sides opposite to congruent angles are congruent in an isoceles triangle. In $\triangle BGC$, $\angle BGC = d$, $\S 81$) angle sum in a triangle is $2d$. $\triangle BGC$ is half the required square.

Use the method of translation ($\S 101$) to translate line $AC$ to be through $G$ and parallel to $AC$; this new line intersects $AF$ at point $Z$. $AZ \parallel BG$ and $ZG \parallel AB$ by construction. $AZGB$ is a parallelogram, $\S 85$) parallel segments cutout by parallel lines are congruent, $\S 86(1)$ a parallelogram has opposite sides congruent. $AB = ZG$.

Use the method in $\S 65$ to erect a perpendicular to $ZG$ at $Z$ and in the direction of $H$; intersection with $CY$ is $H'$. In $\triangle H'ZG$, $\angle H'ZG = d$, and $\angle ZGH' = \frac{1}{2}d$, $\S 77(1)$ parallels transversed corresponding angles are congruent. In $\triangle ZH'G$, $\angle ZH'G = \frac{1}{2}d$, $\S 81$) angle sum of a triangle is $2d$. $\angle ZH'G = \angle ZGH'$, $\triangle ZH'G$ is isoceles by $\S 45(1)$. $ZH' = ZG = AB$.

$\triangle ZH'G = \triangle BGC$, $\S 40(1) SAS$. $H'G = BC$. $H'G$ must be $HG$ as $HG = BC$ by construction.

$\angle CAF = \angle GZJ$, $\S 77(1)$. $\angle CAF = \frac{1}{2}d$ by construction, $\angle GZH = d$, $\angle HZJ = \angle GZJ = \frac{1}{2}d$. $\triangle GZJ = \triangle HZJ$, $\S 40(2)$ ASA. $HJ = GJ$. It follows that $HJ$ is half the given diagonal, $BC$. $HJ$ being the segment of $\angle ACH$ side $CH$ originating at its intersection with the side $AF$ of $\angle CAF$, and the terminal point $H$.

Using the proof above the square can be constructed. Let the given sum of a diagonal and a side be $AB$. Use the method in $\S 63$ to construct an angle at $B$ congruent to $\frac{1}{2}d$ with one side along $AB$ and the other in the plane in the direction of $A$; the latter side is $BC$. Use the method in $\S 63$ to construct an angle at $A$ congruent to $\frac{1}{2}d$ with one side along $AB$ and the other in the direction of $BC$; the latter side intersects $BC$ at point $D$. It was previously shown that $DC$ is congruent to half the diagonal summand of $AB$.

Use a straightedge and compass to construct segment $A'B'$ congruent to $AB$. Set the compass step congruent to $DC$, place the pin at $A'$, and mark a segment $A'Y$ in the direction of $B'$. Place the compass pin at $Y$ and mark a segment $YZ$ in the direction of $B'$. $A'Z = 2(DC)) = diagonal of square$, $ZB'$ must be congruent to the side of the square.

Use the method in $\S 65$ to erect a perpendicular to $ZB'$ at $Z$. Set the compass step congruent to $ZB'$, place the pin at $Z$, and sweep until there is an intersection with this perpendicular; intersection point $M$. Use the method in $\S 65$ to erect a perpendicular to $ZB'$ at $B'$. Place the compass pin at $B'$ and sweep until there is an intersection with this perpendicular; intersection point $P$. Construct line segment $MP$.

In quadrilateral $ZMPB'$, $ZB' = ZM = B'P$, $\angle B'ZM = d = \angle ZB'P$; all by construction. $ZM \parallel B'P$, $\S 71$) two perpendiculars to the same line are parallel. $ZMPB'$ is a parallelogram, $\S 86(2)$ quadrilateral with two opposite sides congruent and parallel. $MP = ZB'$, $\S 85$) opposite sides are congruent in parallelogram. $ZMPB'$ is a square, $\S 92$) parallelogram with all sides congruent and all right angles is a square.

(b)

The adjacent sides of a square are perpendicular, so for any pair of bases the sides are altitudes, $\S 92 + \S 85$) square and altitude definitions.

Let there be a square's diagonal $AC$, side $AE$, and their difference is $EC$. The original square is $ABCD$. Along $AD$ construct a segment congruent to $EC$, the difference, originating at $D$ and in the direction away from $A$, terminal point is $Z$. Use the method in $\S 63$ to construct an angle congruent to $\frac{1}{2}d$ at $Z$ with one side along $AZ$ and the other in the direction of $BC$; the latter side intersects $CD$ at point $Y$, $AC$ at point $X$, and $BC$ at point $W$. Construct segment $CZ$.

$\angle CAD = \frac{1}{2}d$; $\S 92$) squares have all right angles, $\S 91(1)$ rhombus diagonals bisect its angles. $\angle AZW = \frac{1}{2}d$ by construction. $\angle AXZ = d$; in $\triangle AXZ$ $\S 81$) triangle angle sum is $2d$.

$\angle ADC = \angle ZDC = d$, $\S 22$) supplementary angles + $\S 92$) squares have right angles. $\angle DYZ = \frac{1}{2}d$, $\S 81$) triangle angle sum is $2d$. $\angle YXC = d$, $\S 22$) supplementary angles. $\angle XCY = \frac{1}{2}d$, $\S 81$) triangle angle sum.

In $\triangle CAZ$, $AC = AZ$ by construction and this triangle is isoceles, $\S 34$) isoceles triangles. $\angle ACZ = \angle AZC$, $\S 35(2)$ angles at the base of isoceles triangles are congruent. $\angle XCY + \angle YCZ = \angle DZY + \angle YZC$. $\frac{1}{2}d + \angle YCZ = \frac{1}{2}d + \angle YZC$. $\angle YCZ = \angle YZC$. $YC = YZ$, $\S 45(1)$ in a triangle sides opposite to congruent angles are congruent.

$YD = DZ =$ the differece of the diagonal and a side, $\S 45(1)$ sides opposite congruent angles. $YZ$, the hypotenuse of a right triangle whose sides are this difference, is half the side of the original square.

Based on this proof the square can be constructed. Set the compass step congruent to the given difference, $EC$, and construct a segment $E'C'$. Use the method in $\S 65$ to erect a perpendicular to $E'C'$ at $E'$, perpendicular $E'M$. Set the compass pin at $E'$ and mark segment $E'N$ along $E'M$. Construct segment $NC'$. $NC'$ is half the side of the required square.

Set the compass step congruent to $NC'$, place the pin at $N$, and mark a segment $NP$ in the direction away from $C'$. $C'P$ is the side of the square. Use the method in $\S 65$ to erect perpendiculars to $C'P$ at both $C'$ and $P$ on the same side of $C'P$; lines $C'Q$ and $PR$ respectively. Set the compass step congruent to $C'P$, place the pin at $C'$ and mark segment $C'S$ along $C'Q$. Place the pin at $P$ and mark segment $PT$ along $PR$. Construct segment $ST$.

$C'STP$ is the required square; $\S 85$) in parallelogram if on angle is right all are right, $\S 92$) square is a parallelogram with all congruent sides and right angles.

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Forum thread: Methods Of Construction And Symmetries ]]>

Use a compass and straightedge to construct base $A'B'$ congruent to given base $AB$. Use the method in $\S 63$ to construct $\angle A'B'Z$ at point $B'$ with one side along $B'A'$. Mark segment $B'G'$ along $B'Z$ congruent to given lateral $FG$. Use the method in $\S 66$ to drop a perpendicular from $G'$ to $A'B'$, foot is point $H'$. Use the method in $\S 65$ to erect a perpendicular to $G'H$ at point $G'$, a point along the perpendicular is $Y$. $A'B' \parallel YG'$; $\S 71$) two lines perpendicular to the same line are parallel.

Set the compass step congruent to given lateral $DE$, place the pin at $A'$, and sweep an arc which will intersect $YG'$. This intersection could occur in one of two places. If two, chose one of the intersection points (it cannot be one that will cause a line from this point through $A'$ to intersect $B'G'$) and label it $J$.

It is possible that the wrong lateral was chosen to be adjacent to the given angle and this arc does not intersect $YG'$. That would require starting over and choosing this lateral first.

Construct lateral $A'J$. $A'J$ must not be parallel to $B'G'$. $A'B'G'J$ is the required trapezoid; $\S 96$) quadrilateral with one pair of parallel sides and the other sides not parallel is a trapezoid.

(b)

Let there be a trapezoid $WXYZ$ where $YZ$ is the greater base. Mark a segment $YV$ along $YZ$ congruent to base $WX$. Construct line segment $XV$. $\S 96$) in a trapezoid two opposite sides, bases, are parallel. $WX = YV$ by construction. $\S 86(2)$ quadrilateral with two opposite sides parallel and congruent is a parallelogram, $WXYV$. $\S 84 + \S 85$) parallelogram sides are pairwise parallel and congruent. $WY \parallel VX$ and $WY = VX$. Thus, $VZ = YZ - YV$, the difference between the bases. $\triangle VXZ$ has a base that is the trapezoid base difference and sides that are the trapezoid laterals.

Using what was shown about with the triangle, the trapezoid can be constructed. Let the given difference between the bases be $GH$, lateral #1 be $CD$, lateral #2 be $EF$, and diagonal be $AB$. Use the method in $\S 62$ to construct a triangle from $GH$, $CD$, and $EF$ with $GH$ the base, $\triangle G'H'D'$. Extend base $G'H'$ in the direction of $H'$ to a point $M$.

Use the method in $\S 66$ to drop a perpendicular from $D'$ to $MG'$, foot is $N$. Use the method in $\S 65$ to erect a perpendicular to $D'N$ at point $D'$ and into the plane in the direction of $M$; a point along this perpendicular is $P$. $PD' \parallel MG'$, $\S 71$) two perpendiculars to the same line are perpendicular. Set the compass step congruent to diagonal $AB$, place pin at $G'$, and sweep until an intersection with $PD'$; intersection point $O$.

Set the compass step congruent to $OD'$, place ping at $H'$, and mark segment $QH'$ along $MG'$ in the direction of $M$. Construct segment/lateral $OQ$. $OG' = diagonal AB$, $OQ = lateral 2 EF$, $D'G' = laterral 1 CD$, $G'H' = base difference GH$; the required trapezoid is $OD'G'Q$.

(c)

Let the given sides be $AB$ (base #1), $CD$ (base #2), $EF$ (lateral #1), $GH$ (lateral #2). Find the difference of the bases by subtracing the lesser base, $AB$, from the greater, $CD$; the difference is $ZD$. $\#213(b)$ showed that the laterals and the base difference for a triangle within the trapezoid. Use the method in $\S 62$ to construct $\triangle Z'E'D'$ from the given segments $EF$, $GH$, and base difference $ZD$; $ZD$ is the base of the triangle.

Use the method in $\S 66$ to drop a perpendicular to $Z'D'$ from point $E'$, foot is point $N$. Set the compass step congruent to base $CD$, place pin at $D'$, and mark a point $C'$ colinear with $Z'D'$ and in the direction of $Z'$. Construct segment $C'D'$ with straightedge. Use the method in $\S 65$ to erect a perpendicular, $ME'$, to $E'N$ and into the plane in the direction of $C'$. Set the compass step congruent to base $AB$, place pin at $E'$, and mark point $A'$ in the direction of $M$. Construct segment / lateral $A'C'$. $A'E'D'C'$ is the required trapezoid.

This method does not always produce the trapezoid. For example, if the given sides' roles in the trapezoid are not identified, the sides chosen as the bases may not produce the expected outcome. With the given sides above, if the laterals were taken as the bases and bases as the laterals, when constructing the initial triangle the triangle laterals would not intersect when using construction method $\S 62$.

(d)

Let the given base be $AB$, distance between the bases $GH$, diagonal #1 $CD$, diagonal #2 $EF$. Construct a line segment / base congruent to $AB$, $A'B'$. Use the method in $\S 65$ to construct a perpendicular to $A'B'$ at $A'$, perpendicular $A'Z$. Set the compass step congruent to $GH$, place the pin at $A'$, and mark a segment $A'M$ along $A'Z$ in the direction of $Z$.

Use the method in $\S 65$ to construct a perpendicular to $A'M$ at $M$, perpendicular $MN$. $MN \parallel A'B'$, $\S 71$) two lines perpendicular to the same line are parallel. Set the compass step congruent to $CD$, place the pin at $A'$, and sweep until an intersection with line $MN$; intersection point $D'$. Set the compass step congruent to $EF$, place the pin at $B'$, and sweep until an intersection with line $MN$; intersection point $F'$. The required trapezoid is $A'D'F'B'$.

This is possible when the role of the given sides is known or they are accidentally chosen correctly (the distance between bases must be less then or equal to the smaller of the laterals; $\S 51$) perpendicular from a point is the shortest segment from this point to the line.)

(e)

I have not found a solution.

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$\S 92$) square is a rectangle rhombus. $\S 90(1)$ rectangle diagonals are congruent. $\S 91$) rhombus diagonals are perpendicular and bisect each other.

Use a compass and straightedge to construct diagonal segment $AB$ congruent to the given diagonal. Use the method in $\S 67$ to bisect $AB$, midpoint $C$. Use the method in $\S 65$ to erect a perpendicular to $AB$ at $C$ and extend it into the plane on both sides of $AB$. Mark segments $CD$ and $CE$ congruent to $AC$ originating at $C$ and along both directions of this perpendicular.

Construct sides $AD$, $BD$, $BE$, $AE$. $\angle ACD = \angle BCD = \angle BCE = \angle ACE = d$ by construction; $\S 22$) supplementary angles. $\triangle ACD = \triangle BCD = \triangle BCE = \triangle ACE$; $\S 40(1)$ SAS. $ADBE$ is the required rhombus.

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(a)

All sides are congruent to the given side, $\S 91$) rhombus is a parallelogram with all congruent sides, diagonals bisect the angles and are perpendicular to each other. Use a compass and straightedge to construct a diagonal segment $AB$ congruent to the given diagonal. Use the method in $\S 67$ to bisect $AB$, midpoint $M$. Use the method in $\S 65$ to erect a perpendicular to $AB$ at $M$, perpendicular $CD$. Extend $CD$ into both sides of the plane about $AB$.

Set the compass step congruent to the given side, place the pin at $A$, and sweep an arc. This arc intersects $CD$ at two points, $W$ and $X$. Construct line segments $AW$, $WB$, $BX$, and $AX$. $AWBX$ is the required rhombus.

(b)

Construct a diagonal segment congruent to the first given diagonal, $AB$, using a compass and straightedge. Use the method in $\S 67$ to bisect $AB$, midpoint is $M$. Also bisect the second given diagonal, $CD$, midpoint is $N$. Use the method in $\S 65$ to erect a perpendicular to $AB$ at point $M$, perpendicular $MZ$. Extend this perpendicular into both sides of the plane about $AB$.

Mark a segment $MD$ congruent to $CN$ along $MZ$. Mark a segment $ME$ congruent to $CN$ along $MZ$, originating at $M$ and in the direction away from $Z$. Construct segments $AD$, $DB$, $BE$, $AE$. $ADBE$ is the required rhombus, $\S 91$) rhombus is a parallelogram with perpendicular diagonals that bisect each other.

(c)

Construct a line segment $AB$. Use the method in $\S 67$ to bisect $AB$, midpoint $C$. Use the method in $\S 65$ to erect a perpendicular to $AB$ at point $A$, perpendicular $AC$. Mark a segment $AD$ congruent to the given distance between sides along $AC$. Use the method in $\S 65$ to erect a perpendicular to $AD$ at point $D$, perpendicular $DE$. $AB \parallel DE$ and two sides of the rhombus are along these lines. $\S 51$) distance from point to a line is a perpendicular, $\S 71$) two perpendiculars to the same line are parallel.

Set the compass step congruent to the given diagonal, place the pin at point $A$, and sweep an arc until an intersection with $DE$, point $F$. Construct diagonal segment $AF$.

Use the method in $\S 67$ to bisect $AF$, midpoint $G$. Use the method in $\S 65$ to erect a perpendicular to $AF$ at point $G$, perpendicular $GH$. Extend $GH$ on both sides of $AF$ until it intersects $DE$, intersection point $J$, and intersects $AB$, intersection point $K$. Construct segments $AJ$ and $FK$. $AJFK$ is the required rhombus, $\S 91$) rhombus diagonals are perpendicular and bisect each other.

(d)

Construct line segment $AB$. Use the method in $\S 63$ to construct an angle congruent to the given angle at $A$ with one side along $AB$ and the other side with a point $C$. Use the method in $\S 64$ to bisect $\angle ABC$, bisector $AZ$. Mark a segment $AD$ congruent to the given diagonal along $AZ$.

Use the method in $\S 67$ to bisect $AD$, midpoint $E$. Use the method in $\S 65$ to erect a perpendicular to $AD$ at point $E$, perpendicular $EF$. Extend $EF$ into the plane on both sides of $AD$ until it intersects $AB$, intersection point $G$, and intersects $AC$, intersection point $H$. Construct line segments $DG$ and $DH$. $AGDH$ is the required rhombus; $\S 91(1)$ rhombus diagonals bisect the angles of the rhombus and bisect each other.

(e)

Use the method in $\S 63$ to construct $\angle ABC$ congruent to the given angle. Continue side $AB$ through the vertex to a point $D$. $\angle ABC + \angle CBD = 2d$, $\S 22$) supplementary angles. In the rhombus to be constructed, $BC$ is a transversal to parallels $AB$ and the side opposite to it. $\angle CBD$ is the other angle in this rhombus, $\S 77(3)$ parallels transversed same side interior angles sum is 2d.

Use the method in $\S 64$ to bisect $\angle CBD$, bisector is $BE$. Set the compass step congruent to the given diagonal, place the pin at $B$, and mark a point $F$ along $BE$. $\S 91$) rhombus diagonals bisect its angles. Use the method in $\S 67$ to bisect diagonal $BF$, midpoint $G$. Use the method in $\S 65$ to erect a perpendicular to $BF$ at point $G$ and in the direction of $D$, perpendicular $GH$. Extend this perpendicular into the other side of the plane toward $C$, its intersection with $BC$ is $J$.

Construct sides $JF$ and $HF$. Diagonals $BF$ and $HJ$ are perpendicular by construction. All the angles about $G$ are right, $\S 22$) supplementary angles. $\angle CBF = \angle DBF$; bisector construction. $\triangle DBG = \triangle CBG$; $\S 40(2)$ ASA. $\triangle DBG = \triangle DFG$; $\S 40(1)$ SAS. $\triangle CBG = \triangle CFG$; $\S 40(1)$ SAS.

Therefore all of these triangles are congruent and the sides of the quadrilateral are congruent. $BDFC$ is the required rhombus; $\S 86(1)$ quadrilateral with congruent opposite sides, $\S 91$) parallelogram with all congruent sides is a rhombus.

(f)

Use a compass and a straightedge to construct diagonal $AB$ congruent to the given diagonal. Use the method in $\S 63$ to construct $\angle CAB$ at point $A$ congruent to the given angle with one side along $AB$. Again do this to construct $\angle DAB$ at point $A$ congruent to $\angle CAB$ with one side along $AB$ and the other into the side of the plane that does not contain $C$. Use this method again to construct $\angle MBA$ at point $B$ congruent to $\angle CAB$ with one side along $BA$. Sides $BM$ and $AC$ intersect at point $Y$.

Use the same method to construct $\angle NBA$ at point $B$ congruent to $\angle CAB$ with one side along $BA$ and the other into the part of the plane that does not contain $M$. Sides $BN$ and $AD$ intersect at point $Z$. $\S 85$) opposite angles in a parallelogram are congruent, $\S 91(1)$ rhombus diagonals bisect the angles.

$\angle YAB = \angle ZAB = \angle YBA = \angle ZBA$ by construction. $\triangle YAB = \triangle ZAB$; $\S 40(2)$ ASA and these are isoceles $\S 33$. $YA = YB = ZA = ZB$; $\S 45(2)$ in a triangle sides opposite to congruent angles are congruent. $YAZB$ is the required rhombus; $\S 86(1)$ quadrilateral with opposite sides congruent is a parallelogram, $\S 91$) parallelogram with all congruent sides is a rhombus.

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Forum thread: Methods Of Construction And Symmetries ]]>

The unknown diagonal is congruent to the given diagonal, $\S 90$. This is the same problem as #209(c).

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(a)

Use a compass and a straightedge to construct a side $C'D'$ congruent to given side $CD$. Set the compass step congruent to given diagonal $EF$, place the pin at $C'$, and mark an arc. Set the compass step congruent to given side $AB$, place the pin at $D'$, and mark an arc. The intersection point of the previous two arcs is $B'$. Connect $D'$ and $B'$ by line.

Use the method in $\S 63$ to construct an angle congruent to $\angle C'D'B'$ at point $B'$ with one side along $D'B'$ in the direction opposite $D'$, and the other side in the same part of the plane as $C'D'$. The latter angle side is $B'M$. Mark a segment congruent to $C'D'$ along $B'M$, originating at $B'$ and in the direction of $M$, terminal point $N$. Construct line $C'N$.

$\angle C'D'B' = \angle NB'Z$, $C'D' \parallel NB'$; $\S 73(1)$ two lines with congruent corresponding angles on a transverse are parallel. $C'D' = NB'$ by construction. $C'D'B'N$ is a parallelogram, $\S 86(2)$ quadrilateral with two opposite sides parallel and congruent is a parallelogram.

(b)

Use a compass and a straightedge to construct a side $A'B'$ congruent to the given side. $\S 87(1)$ parallelogram diagonals bisect each other. Use the method in $\S 67$ to bisect the given diagonals $CD$ and $EF$, midpoints $G$ and $H$ respectively. Set the compass step congruent to $CG$, place the pin at $A'$, and mark an arc. Set the compass step congruent to $EH$, place the pin at $B'$, and mark an arc. The intersection point of these two arcs, $Z$, is the intersection point of the diagonals in the parallelogram.

Construct a segment congruent to $CD$, originating at $A'$, and through $Z$; terminal point $D'$. Construct a segment congruent to $EF$, originating at $B'$, and through $Z$; terminal point $F'$.

Construct line segments $D'F'$, $B'D'$, $A'F'$. $A'B'D'F'$ is the required parallelogram.diagonals in the parallelogram.

Construct a segment congruent to $CD$, originating at $A'$, and through $Z$; terminal point $D'$. Construct a segment congruent to $EF$, originating at $B'$, and through $Z$; terminal point $F'$.

Construct line segments $D'F'$, $B'D'$, $A'F'$. $A'B'D'F'$ is the required parallelogram.

(c)

Use a compass and straightedge to construct a segment $C'D'$ congruent to given diagonal $CD$. Use the method in $\S 67$ to bisect this segment, the midpoint is $Z$. Also bisect given diagonal $EF$, its midpoint is $Y$.

Use the method in $\S 63$ to construct an angle congruent to the given angle at point $Z$ with one side along $ZD$. The other side is $ZX$. Extend $ZX$ into the plane on both sides of $CD$. Mark segment $ZV$ congruent to $EY$ along side $ZX$. Mark segment $ZW$ congruent to $EY$ along side $ZW$. $\S 87(1)$ parallelogram diagonals bisect each other.

Construct segments $C'W$, $WD$, $C'V$, $DV$. $C'WDV$ is the required parallelogram.

(d)

Use a compass and straightedge to construct side $A'B'$ congruent to given side $AB$. Use the method in $\S 65$ to erect a perpendicular at $A'$. Mark a segment $A'D'$ congruent to given altitude $CD$ along this perpendicular. Use the method in $\S 65$ to erect a perpendicular at $D'$ into the area of the plane containing $B'$, perpendicular $D'M$.

Set the compass step congruent to given diagonal $EF$, place the pin at $A'$, and sweep until it intersects $D'M$; intersection point is $Z'$. Mark a segment $Z'Y'$ congruent to $AB$ along $Z'D'$. $\S 71$) two lines perpendicular to the same line are parallel. $\S 84 + \S 85$) parallelogram has pairwise parallel and congruent sides. $A'B' = Y'Z'$ by construction, $A'B' \parallel Y'Z'$ by construction.

Construct segments $A'Y'$ and $B'Z'$. $A'Y'Z'B'$ is the required parallelogram, $\S 86(2)$ quadrilateral with two opposite sides parallel and congruent is a parallelogram.

This construction assumed the altitude was from a parallelogram where the given side is one of the bases. If the altitude were to the unknown sides, it seems that it might be impossible to determine the angle between the given side and its adjacent sides.

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Forum thread: Methods Of Construction And Symmetries ]]>

Use a compass and straightedge to construct a segment of the quadrilateral congruent to the largest given segment, $C'D'$. Set the compass step congruent to given side $AB$, place the pin at $D'$, and mark an arc. Set the compass step congruent to given diagonal $BD$, place the pin at $C'$, and mark an arc. The intersection of these two arcs is point $Z$.

Set the compass step congruent to given side $AD$, place the pin at $C'$, and mark an arc. Set the compass step congruent to given diagonal $AC$, place pin at $D'$, and mark an arc. The intersection of these two arcs is point $Y$.

Construct line segments $C'Y$, $YZ$, and $D'Z$. $C'YZD'$ is a quadrilateral, $C'D' = CD$, $C'Y = AD$, $YZ = BC$, $D'Z = AB$, $C'Z = BD$, $D'Y = AC$.

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(a)

Use a compass and straightedge to construct segment, $AB$, congruent to the given base. Extend the segment in the direction of $B$, a length congruent to $AB$, terminal point $A'$. At an arbitrary point $D$ along $AA'$ erect perpendicular $CD$ using the method in $\S 65$. Use the method in $\S 65$ to erect a perpendicular to $CD$ at $C$ and extend it in both directions, line $YZ$. $YZ$ and $AA'$ are both perpendicular to $CD$ and parallel, $\S 71$) two perpendiculars to the same line are parallel.

Set the compass step congruent to the given lateral side, place the pin at $B$, and sweep until it intersects $YZ$ twice, points $F$ and $F'$. Connect $B$ and $F$ by line, and $B$ and $F'$ by line, $BF = BF' = given lateral$.

$\triangle FBF'$ is isoceles, $\S 33$) isoceles triangle has two congruent sides, and $\angle BFF' = \angle BF'F$, $\S 35(2)$ base angles are congruent in an isoceles triangle.

$\angle BF'F = \angle F'BA'$ and $\angle BFF' = \angle FBA$, $\S 77(1)$ parallel lines corresponding angles are congruent.

$\angle F'BA' = \angle FBA$. $\triangle AFB = \triangle A'F'B$, $\S 40(1)$ SAS.

$\triangle AFB$ or $\triangle A'F'B$ is the required triangle.

(b)

Use a compass and straigtedge to construct the base segment, $AB$, congruent to the given base. Use the method in $\S 65$ to erect a perpendicular at the midpoint of $AB$, perpendicular $CD$. Use the method in $\S 65$ to erect a perpendicular to $CD$ at point $C$, perpendicular $YZ$. $YZ \parallel AB$, $\S 71$) two perpendiculars to the same line are parallel.

Use the method in $\S 63$ to construct an angle congruent to the given angle at point $A$ with one side along $AB$ and the other side on the same side of $AB$ as $YZ$; the latter side's intersection with $YZ$ is point $E$.

Construct line $EB$. $\triangle EAB$ is the required triangle.

(c)

Let the given altitudes be $CD$ and $EF$. Construct a line segment $AZ$. Use the method in $\S 63$ to construct an angle congruent to the given angle at point $A$ with one side along $AZ$ and the remaining side is $AM$. Use the method in $\S 65$ to erect a perpendicular congruent to $CD$ at point $A$ and in the same part of the place as $AM$, perpendicular $AC'$. Use the same method to erect a perpendicular to $AM$ at point $M$, congruent to $EF$, and in the same part of the plane as $AZ$, perpendicular $ME'$.

Use the method in $\S 65$ to erect a perpendicular to $AC'$ at $C'$, perpendicular $C'H$, and erect a perpendicular to $E'M$ at $E'$, perpendicular $E'J$. The intersection point of these perpendiculars if $G$. $C'H$ intersects $AM$ at point $K$. $E'J$ intersects $AZ$ at $L$. Construct line $KL$. $\triangle AKL$ is the required triangle.

(d)

Use a compass and straightedge to construct a segment congruent to the given sum of the other two sides, $AB$. Use the method in $\S 65$ to erect a perpendicular to $A$. Mark a segment congruent to the given altitude, originating at $A$, along this perpendicular, segment $AC$. Again use the method in $\S 65$ to erect a perpendicular to $AC$ at $C$ and in the direction of $B$, perpendicular $CZ$.

Set the compass step congruent to the given side, place the pin at $A$, and sweep until it intersects $CZ$, intersection point $D$. Construct line $DB$.

Use the method in $\S 63$ to construct an angle at $D$ congruent to $\angle DBA$ with one leg along $DB$ and the other in the direction of $AB$. The latter intersects $AB$ at point $E$. $DE = BE$, $\S 45(1)$ sides opposite to congruent angles are congruent. $\triangle ADE$ is the required triangle.

(e)

Use a compass and straightedge to construct a segment $AB$ congruent to the given perimeter. Use the method in $\S 65$ to erect a perpendicular to $AB$ at $A$. Mark a segment along this perpendicular congruent to the altitude, originating at $A$; segment $AC$. Use the method in $\S 65$ to erect a perpendicular to $AC$ at $C$ and in the direction of $B$, $CZ$. Use the method in $\S 63$ to construct an angle at $A$ congruent to the given angle with a side along $AB$ and the other in the direction or $CZ$. The latter intersects $CZ$ at point $D$. $AD$ is a side of the triangle.

Subtract side $AD$ from the perimeter segment by marking a segment along $AB$ originating at $B$ and in the direction of $A$, terminal point $E$. Construct line $DE$.

Use the method in $\S 63$ to construct an angle congruent to $\angle DEF$ at point $D$ with one side along $DE$ and the other in the direction of $AB$. The latter intersects $AB$ at point $F$. $DF = EF$, $\S 45(1)$ sides opposite to congruent angles are congruent. $\triangle ADF$ is the required triangle.

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Forum thread: Methods Of Construction And Symmetries ]]>

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Forum thread: Methods Of Construction And Symmetries ]]>

Let the given angle be $\angle ABC$ and the given segment be $MN$.

Use the method in $\S 64$ to bisect $\angle ABC$, the bisector is $BD$.

Use the method in $\S 67$ to bisect the segment $MN$, midpoint $O$.

Use the method in $\S 65$ to erect a perpendicular to $BD$ at $B$, extend it into the plane on both sides of $BD$.

Use a compass to mark a segment congruent to $MO$ originating at $O$ and in the direction of the plane containing $A$, terminal point $P$.

Make another segment of the same originating at $O$, but in the opposite direction, terminal point $Q$.

Erect a perpendicular at $P$ in the direction of $D$, intersection with side $AB$ is $R$.

Erect a perpendicular at $Q$ in the direction of $D$, intersection with $CB$ is $S$. Connect $R$ and $S$ with a line.

$PR \parallel QS$, $\S 71$) two perpendiculars to the same line are parallel.

$\angle PBR = d - \angle ABD$; $\angle ABD = \angle CBD$; $\angle QBS = d - \angle CBD$; all by construction. $\angle PBR = \angle QBS$.

$\angle BPR = d = \angle BQS$, by construction. $BP = BQ$, by construction. $\triangle BPR = \triangle BQS$, $\S 40(2)$ ASA.

$BR = BS$ and $PR = QS$. $QPRS$ is a parallelogram, $\S 86(2)$ two opposite congruent and parallel sides.

$MN = PQ = RS$, $\S 85$) parallelogram opposite sides are congruent.

$RS$ is the required segment between sides, and $BR$ and $BS$ are the congruent cut-off segments along each side.

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Let the given angle be $\angle ABC$, given line intersecting it be $DE$, and the given segment to be matched be $MN$. $DE$ intersects $AB$ at point $Z$ and side $BC$ at point $Y$.

Use a compass to mark a segment congruent to $MN$ originating at point $Y$ and in the direction of $D$, terminal point $W$.

Use the method in $\S 63$ to construct an angle congruent to $\angle BYD$ at point $W$ with one side along $DE$ and away from $E$, and the other side in the plane toward $B$; a point along the latter side is $X$.

Extend $WX$ in whichever direction has it intersect side $AB$, this is point $R$.

$WX \parallel CB$ by construction, $\S 73(1)$ two lines transversed with congruent corresponding angles.

Use a compass to mark a segment along $CB$ congruent to $WR$, originating at $Y$, and in the direction of the plane containing $R$; terminal point $S$.

Connect $R$ and $S$ by a line. $WR$ and $YS$ are congruent by construction and parallel.

$SRWY$ is a parallelogram, $\S 86(2)$ quadrilateral with two opposite sides that are congruent and parallel.

$SR = YW = MN$, $\S 85$) parallelogram opposite sides are congruent.

$SR$ is the required segment.

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Forum thread: Parallelograms And Trapezoids ]]>

Let the given angle be $\angle ABC$ and given segment $DE$. Using the method in $\S 65$ erect a perpendicular at $B$ perpendicular to $BC$ and in the direction of $A$.

Make a segment on this perpendicular originating at $B$ and congruent to $DE$, terminal point is $F$.

Use the method in $\S 65$ to erect a perpendicular at $F$ and in the direction of $BA$, intersection with the latter is point $G$.

Use the method in $\S 66$ to drop a perpendicular from $G$ to $BC$, foot point is $H$.

$FG \parallel BH$, $\S 71$) two perpendiculars to the same line are parallel.

$DF = FB$ by construction. $GH = FB$, $\S 85$) parallel lines are everywhere the same distance apart.

$GH = DE$, $GH$ is between the sides, and perpendicular to side $BC$.

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From the given point drop a perpendicular to the closest given parallel line using the method in $\S 66$. Next erect a perpendicular at the given point perpendicular to the perpendicular just constructed. This latest perpendicular is parallel to the given parallels, $\S 71$) two perpendiculars to the same line are parallel.

Set the compass step congruent to the given segment that is to be matched. Place the pin along the parallel that the perpendicular was dropped to in the first step, and sweep the compass until it intersects the other given parallel.

Connect by line the pin point and the point of intersection with the parallel. This newly constructed segment is congruent to the given segment to be matched.

Use the method in $\S 63$ to construct an angle, with a side along the third parallel, congruent to the angle the segment just constructed makes with the given parallel.

Extend the side of this new angle not along the third parallel to intersect the two given parallels, $\S 73(1)$ this extended side and angle the constructed given segment match would make with the third parallel are corresponding and congruent by construction.

Thus, the extended angle side and constructed given segment match are parallel.

$\S 84$) sides of a parallelogram are pairwise parallel and $\S 85$ parallel segments cut out by parallel lines are congruent.

The segment of the extended angle side between the given parallels is congruent to the given segment to match.

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Forum thread: Parallelograms And Trapezoids ]]>

I must not understand, the location of the point is not specified. I don't believe it can be located randomly between the two given lines, as I expect there is a proof that shows at a certain point the line segment through the point is not bisected at the point.

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Forum thread: Parallelograms And Trapezoids ]]>

$\S 51$) distance from a point to a line is the perpendicular dropped from the point to the line.

Use the method in $\S 65$ to erect a perpendicular at a point on the given line. Mark a segment on this perpendicular congruent to the given distance with a compass, one end of the segment is the foot of the perpendicular.

At the point of this new segment, not the foot, erect a perpendicular to it using the method in $\S 65$. This is the desired parallel, $\S 71$) two perpendiculars to the same line are parallel and $\S 85$) parallel lines are everywhere the same distance apart.

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Forum thread: Parallelograms And Trapezoids ]]>

I don't understand what is to be shown.

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Forum thread: Parallelograms And Trapezoids ]]>

$\S 51$) distance from point to a line is the perpendicular dropped from the point to the line.

$\S 85$) If two lines are parallel, then all points of each of them are the same distance away from the other line.

Erect two perpendiculars between the two given lines. Connect the midpoint of these perpendiculars with a line. All points along this line connecting the midpoints are equidistant from the two given lines (half the distance between them).

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Forum thread: Parallelograms And Trapezoids ]]>

Let the given point be $M$, given line $AB$, with segments between them: $MC$, $MF$.

Construct a trapezoid by constructing a line segment through point $M$, parallel to $AB$, and less than $AB$; $XY$.

Connect lateral sides $AX$ and $BY$. Connect the midpoints of sides $AX$ and $BY$, this is the trapezoid midline.

The points of intersection of the midline with the segments from $M$ to line $AB$ are the midpoints, #184) trapezoid midline bisects any segment connecting the bases of the trapezoid.

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Forum thread: Parallelograms And Trapezoids ]]>

$\S 92$) a square is a rhombus with all right angles. $\S 91$) a rhombus is a parallelogram with all congruent sides.

$AA' = BB'$, $\angle A'AD' = \angle A'BB' = d$, $AD' = A'B$; all given. $\triangle A'AD' = \triangle A'BB'$, $\S 40(1)$ SAS.

$A'B' = A'D' = B'C' = C'D'$, $\S 85$) parallelogram sides are pairwise congruent.

$\angle A'B'B + \angle A'B'C' + \angle C'B'C = 2d$, $\S 22$) supplementary angles.

In $\triangle CB'C'$, $\angle C'B'C + \angle B'CC' + \angle B'C'C = 2d$, $\S 81$) triangle angle sum is $2d$. $\angle B'CC' = d$, given; $\angle C'B'C + \angle B'C'C = d$.

$\triangle A'BB' = \triangle B'CC'$, $\S 40(3)$ SSS. $\angle A'B'B = \angle B'C'C$. From this congruence, $\angle C'B'C + \angle A'B'B = d$.

$\angle A'B'C' = d$. All the angles in rhombus $A'B'C'D'$ are right, $\S 85$) if one angle in a parallelogram is right, they all are right.

Quadrilateral $A'B'C'D'$ is a square, $\S 91$ and $\S 92$.

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Forum thread: Parallelograms And Trapezoids ]]>

$\S 92$) Square is rhombus with all right angles. $\S 91$) rhombus sides are all congruent.

$AC' = CA'$, given halves of congruent sides. $AC' \parallel CA'$, given square parallelogram.

Quadrilateral $AC'CA'$ has two congruent and parallel sides, so it is a parallelogram, $\S 86$. $C'C \parallel AA'$.

The same method is used to show quadrilateral $BB'DD'$ is a parallelogram and $BB' \parallel D'D$.

$C'C$ intersects $BB'$ at $W$ and $DD'$ at $X$. $AA'$ intersects $BB'$ at $Y$ and $DD'$ at $Z$.

$\S 85$) parallel lines cut out parallel segments, $WY = XZ$ and $WX = YZ$.

Quadrilateral $WXZY$ is a parallelogram, $\S 86$.

At point $C'$ construct a parallel to $BB'$, intersection with $AA'$ is $V$. $C'W = VY$, $\S 85$.

$\angle WBC' = \angle VC'A$, $\S 77(1)$ parallel lines transversed corresponding angles. $BC' = C'A$, given.

$BW = WY$, $\S 93$) $\angle ABB'$ parallels through congruent segments along one side of the angle cut out congruent segments along the other side.

$\triangle WBC' = \triangle VC'A$, $\S 40(1)$, SAS. Thus, $C'W = AV = VY$. $V$ is the midpoint of $AY$.

$AB = AD$, $\angle BAB' = \angle ADA' = d$, $AB' = DA'$. All given. $\triangle ABB' = \triangle DAA'$, $\S 55(2)$ right triangle congruence by hypotenuse and leg.

$\angle DAA' = \angle ABB'$. $\angle ABB' + \angle B'BD' = d$. $\angle B'BD' = \angle DD'C$, $\S 77(2)$ alternate angles. $\angle DD'C = \angle ADZ$, $\S 77(2)$.

In $\triangle DAA'$, $\angle ADZ + \angle DAA' = d$, $\angle AZD$ must be congruent to $d$, $\S 81$; triangle angle sum is $2d$.

$\angle XZY$ supplementary to $\angle AZD$ must also be right, $\S 22$. All angles in parallelogram $WXZY$ are right, $\S 85$.

$\angle BYA$ supplementary to $\angle WYZ$ is right, $\S 22$.

$AB = AD$, $\angle ABY = \angle DAZ$, $\angle BYA = \angle AZD = d$. $\triangle ABY = \triangle DAZ$, #158 ) SAA.

$AB = B'D$, given. $AY = YZ$, $\S 93$) $\angle DAA'$ parallels $B'B$ and $DD'$ cut-off congruent segments along side $AA'$.

Segment $BY$ has congruent summands $BW + WY$ and is congruent to segment $AZ$ with congruent summands $AY + YZ$. $BW = WY = AY = YZ$.

All sides of parallelogram $WXZY$ are congruent with all right angles. This is a square, $\S 92$.

$\angle ADZ + \angle DAA' = d = \angle DAA' + \angle C'AY$. $\angle C'VA = d = \angle BYA$, $\S 77(1)$ corresponding angles.

$\angle ADC = d = \angle ADZ + \angle A'DZ = \angle ADZ + \angle DAA'$.

$\angle A'DZ = \angle DAA'$. $\angle A'ZD = \angle XZY = d$, $\S 26$) veritcal angles. $AC' = DA'$, given.

$\triangle AC'V = \triangle DA'Z$, #158) SAA. $A'Z = AV$.

$AA' = AY + YZ + ZA' = AV + VY + YZ + ZA'$. $YZ$ was shown congruent to $AY$, thus it is congruent to $2(AV)$.

$AA' = 5(AV)$ and side $YZ$ of the quadrilateral is 2 of these summands.

$BB' = AA'$ and $WY = YZ$. $WY'$ is $\frac{2}{5}$ of $BB'$.

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