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Exercise 205: Between the sides of a given angle, place a segment congruent to a given segment and such that it cuts congruent segments on the sides of the angle.

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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Exercise 205 by blakesterblakester, 17 Sep 2017 23:18

Exercise 204: Between the sides of a given angle, place a segment congruent to a given segment and parallel to a given line intersecting the sides of the angle.

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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Exercise 204 by blakesterblakester, 17 Sep 2017 23:17

Exercise 203: Between the sides of a given angle, place a segment congruent to a given segment and perpendicular to one of the sides of the angle.

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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Exercise 203 by blakesterblakester, 17 Sep 2017 23:16

Exercise 202: Through a given point, draw a line such that its line segment, contained between two given parallel lines, is congruent to a given segment.

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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Exercise 202 by blakesterblakester, 17 Sep 2017 23:15

Exercise 201: Through a given point, draw a line such that its line segment, contained between two given parallel lines, is congruent to a given segment.

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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Exercise 201 by blakesterblakester, 17 Sep 2017 23:14

Exercise 200: Draw a line parallel to a given one and situated at a given distance from it.

$\S 51$) 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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Exercise 200 by blakesterblakester, 17 Sep 2017 23:14

Exercise 199: The vertices of triangles having a common base and congruent altitudes. (geometric locus)

I don't understand what is to be shown.

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Exercise 199 by blakesterblakester, 17 Sep 2017 23:13

Exercise 198: The points equidistant from two given parallel lines. (gemometric locus)

$\S 51$) 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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Exercise 198 by blakesterblakester, 17 Sep 2017 23:13

Exercise 197: Find the midpoints of all segments drawn from a given point to various points of a given line. (geometric locus)

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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Exercise 197 by blakesterblakester, 17 Sep 2017 23:12

Exercise 196: Given a square ABCD. On its sides, congruent segments AA', BB', CC', and DD' marked. The points A', B', C', and D' are connected consecutively by lines. Prove that A'B'C'D' is a square.

$\S 92$) 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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Exercise 196 by blakesterblakester, 17 Sep 2017 23:12

Exercise 195: Let A', B', C', and D' be the midpoints of the sides CD, DA, AB, and BC of a square. Prove that the segments AA', CC', DD', and BB' cut out a square, whose sides are congruent to 2/5th of any of the segments.

$\S 92$) 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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Exercise 195 by blakesterblakester, 17 Sep 2017 23:09

Exercise 194: Bisectors of the angles of a rectangle cut out a square.

Let the rectangle be $ABCD$, bisector of $\angle A$ is $AZ$, $\angle B$ is $BY$, $\angle C$ is $CX$, $\angle D$ is $DW$.
$\S 90$) in a rectangle all angles are right. $\angle A = \angle B = \angle C = \angle D = d$.
The intersection of bisectors $AZ$ and $BY$ is $M$. Drop altitude from $M$ to $AB$, foot is point $N$. $\triangle MNA = \triangle MNB$, #158) SAA.

$BC \parallel AD$, given. $\angle BZA = \angle DAZ$, $\S 77(1)$ parallel transverse corresponding angles. $\angle BSA = \frac{d}{2}$.
$\angle BCX = \frac{d}{2}$, given. $AZ \parallel XC$, $\S 73(1)$ two lines transversed and corresponding angles are congruent.

$\angle CWD = \angle ADW$, $\S 77(1)$. $\angle CWD = \frac{d}{2}$. $\angle CBY = \frac{d}{2}$, given. $BY \parallel WD$, $\S 73(1)$.
The sides of the cut-out quadrilateral are pairwise congruent, $\S 85$) parallel segments cut out by parallel lines are congruent.
$\angle AMB = d$, $\S 81$) angle sum of triangle is $2d$.
$\angle AMB$ and $\angle YMZ$ are vertical, $\S 26$, so $\angle YMZ = d$.
$\triangle AMB = \triangle AMY$, $\S 40(2)$, ASA; $MB = MY$.

$NM$ is the midline in $\triangle ABY$ and parallel to $AD$. Intersection with $CD$ is point $O$. $AN = DO$, $\S 85$.
Altitude in $\triangle CPD$ can be shown to have foot at $O$, so $PO$ is along $NO$.
$MP$ is diagonal in the cut-out rectangle. $\angle ZMO = \angle NMA$, $\S 26$) vertical angles.
$\angle YMO = \angle NMB$, $\S 26$. But $\angle NMB = \angle NMA$, so $\angle ZMO = \angle YMO$.
#192) a quadrilateral with angle bisected by diagonal is a rhombus. All sides of the cut-out quadrilateral are congruent and all angles are right, so it is a square, $\S 92$.

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Exercise 194 by blakesterblakester, 17 Sep 2017 15:06

Exercise 193: From the intersection point of the diagonals of a rhombus, perpendiculars are dropped to the sides of the rhombus. Prove that the feet of these perpendiculars are vertices of a rectangle.

Let the rhombus be $ABCD$ and intersection point of the diagonals is $E$. From point $E$ drop perpendiculars to the four sides: to side $AB$ foot $G$, to side $BC$ foot $F$, to side $CD$ foot $J$, and to side $AD$ foot $H$.
$\angle A = \angle C$ and $\angle B = \angle D$, $\S 85$) opposite angles in a parallelogram are congruent.
$\angle FCE = \angle JCE = \angle HAE = \angle GAE$, $\S 91(1)$ Rhombus diagonals bisect vertex angles.
Right triangles $\triangle CFE = \triangle CJE = \triangle AGE = \triangle AHE$, $\S 55(1)$ acute angle and hypotenuse congruent in right triangles. Thus, $CF = CJ = AG = AH$.
$\triangle GAH = \triangle FCJ$, they are isoceles $\S 33$, $CE$ and $AE$ are altitudes to their bases; $\S 35(1)$ bisectors are altitudes and medians in isoceles triangles.

$FJ$ and $GH$ are perpendicular to diagonal $AC$ and congruent. $GH \parallel FJ$, $\S 71$) two lines perpendicular to the same line are parallel.
$GFJH$ is a parallelogram and $GF \parallel HJ$, $GF = HJ$; $\S 86$) convex quadrilateral with two opposite sides congruent and parallel is a parallelogram.

Supplementary angle ($\S 22$) $\angle AC = \angle AEG + \angle GEB + \angle CEB = \angle AEG + \angle GEB + d$ also $\angle AC = \angle CEJ + \angle JED + \angle AED = \angle CEJ + \angle JED + d$. It was previously shown $\angle AEG = \angle CEJ$. Therefore $\angle GEB = \angle JED$.
$\angle AEG$ and $\angle CEJ$ have common side along $AC$ and are both supplementary to $\angle GEB + d$. Their remaining sides $GE$ and $JE$ must be continuations of one another and they are vertical angles, $\S 26$.

In $\triangle HGJ$, $GH$ is bisected by $AC$, $GJ$ is bisected by $AC$. $AC \parallel HJ$, $\S 95$) triangle midline is parallel to the third side. It was shown $AC \perp FJ$. $\S 77(3)$ parallel lines transversed by another line same side angles sum to $2d$, $\angle FJH$ must be congruent to $d$.

$\S 85$) if one angle in a parallelogram is right then all the angles are right. $\S 90$) a parallelogram with all right angles is a rectangle. $GFJH$ is a rectangle and the feet of the given perpendiculars are at the vertices.

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Exercise 193 by blakesterblakester, 17 Sep 2017 15:05

Exercise 192: Any parallelogram whose angle is bisected by the diagonal is a rhombus.

Let the parallelogram be $ABCD$, diagonal $BD$, and $\angle ABD = \angle CBD$.
$\angle ABD = \angle CDB$, $\S 77(2)$ parallel transverse alternate angles are congruent.
$\angle ABC = \angle ADC$, $\S 85$) parallelogram opposite angles are congruent. $\angle ABC = \angle ABD = \angle CBD$, but these summands are congruent so $\angle ABC = (2)\angle ABD$.
$\angle ADC = \angle CDB + \angle ADB = \angle ABD + \angle ADB$. $\angle ADB$ must be congruent to $\angle ABD$.
$\triangle BCD = \triangle BAD$, $\S 40(2) ASA$. $AB = AD = CD = BC$, $\S 45(1)$ in a triangle sides opposite congruent angles are congruent.

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Exercise 192 by blakesterblakester, 17 Sep 2017 15:04

Exercise 191: A parallelogram whose diagonals are perpendicular to each other is a rhombus.

Let the parallelogram be $ABCD$ with diagonals intersecting at point $E$.
$AE = CE$, $\S 87(1)$ parallelogram diagonals bisect each other. $\angle AEB = \angle CEB$, given perpendicular. $\triangle AEB = \triangle CEB$, $\S 40(1) SAS$. Thus, $AB = BC$.
$AD = BC = AB = CD$, $\S 85$) parallelogram opposite sides are congruent. $\S 91$) a parallelogram with all sides congruent is a rhombus. Parallelogram $ABCD$ is a rhombus.

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Exercise 191 by blakesterblakester, 17 Sep 2017 15:04

Exercise 190: A parallelogram whose diagonals are congruent is a rectangle.

Let the parallelogram be $\S ABCD$ and diagonal intersection point is $E$.
$\S 84$) Parallelogram sides are pairwise parallel. $\S 87(1)$ Parallelogram diagonals bisect each other;
All diagonal segments from point $E$ to the vertices are congruent, given congruent diagonals.
$\angle AEB = \angle DEC$, $\angle BEC = \angle AED$; $\S 26$) vertical angles.
$\triangle BEC = \triangle AED$ and $\triangle AEB = \triangle CED$; $\S 40(1) SAS$.

$\angle BCD = \angle BCE + \angle DCE$ and $\angle ADC = \angle ADE + \angle CDE$. $\angle BCE = \angle ADE$ by the triangle congruent and $\angle DCE = \angle CDE$ by the isoceles triangle.
Both $\angle BCD$ and $\angle ADC$ have the same summands and sum to $2d$, $\S 77(3)$ the sum of same side interior angles on a parallel transverse is $2d$. The measure of one of them must be $d$.
$\S 90$) When one angle of a parallelogram is right, all angles are right and it is a rectangle. Parallelogram $ABCD$ is a rectangle.

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Exercise 190 by blakesterblakester, 17 Sep 2017 15:04

Exercise 189: In an equilateral triangle, the sum of the distances from an interior point to the sides of this triangle does not depend on the point, and is congruent to the altitude of the triangle.

Let the equilateral triangle be $\triangle ABC$, altitude to base $AC$ is $BG$, arbitrary point $M$, distance $M$ to $AB$ is $MD$, distance $M$ to $BC$ is $MY$, and distance $M$ to $AC$ is $MZ$.

$MZ \parallel BG$, $\S 71$) two perpendiculars to the same line are parallel. Through point $M$ construct a line parallel to $AC$, $DE$; its intersection with $BG$ is point $F$. $MZ = FG$, $\S 85$) parallel segments cut-off by parallel lines are congruent.
$BG$ bisects $\angle B$, $\angle DBF = \angle EBF$, $\S 36$) altitude is same as bisector in isoceles triangle. $\angle BFE = \angle FGC = d$, $\S 77(1)$ corresponding angles are congruent in parallels transversed.
$\angle DFB = \angle EFB$, $\S 22$) supplementary angles. Therefore, $\triangle DBF = \triangle EBF$, $\S 40(2)$ ASA.
$\angle BDF = \angle BEF$. $\triangle DBE$ is isoceles, $\S 35$, and point $M$ is along its base $DE$.

Use #187 to show the sum of the remaining distances from $M$, $MX + MY$, is congruent to the remaining segment of the altitude, $BF$.

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Exercise 189 by blakesterblakester, 17 Sep 2017 15:03

Exercise 188: How does this theorem change if points on the extension of the base are taken instead?

Let the given isoceles triangle be $\triangle ABC$. Extend one side of base $AC$ to point $D$. Drop perpendiculars from $D$ to $AB$, $DY$, and $D$ to $BC$, $DW$. These are the distance segments. Construct altitude from $C$ to lateral $AB$, $CZ$. Construct a line through $C$ parallel to $AB$, intersection with $DY$ is point $X$.

$\angle AYX = \angle CXD$, $\S 77(1)$ parallels transversed corresponding angles are congruent.
$\angle BAC = \angle XCD$, $\S 77(1)$.
$\angle BAC = \angle BCA$, $\S 35(2)$ isoceles triangle base angle congruent. $\angle BCA = \angle DCW$, $\S 26$) vertical angle congruence.
It follows that $\angle DCW = \angle DCX$.
$\angle DWC$ is perpendicular to $BC$ as constructed. $\angle DXC$ is perpendicular to $BC$ as shown. $\angle DWC = \angle DXC$.
$\triangle DCX = \triangle DCW$, $\S 55(1)$ right triangle congruent by hypotenuse and acute angle congruence.
$YX = ZC$, $\S 85$) parallel segments cut-off by parallel lines. $YX = DY - DW$.

The theorem changes to: In an isoceles triangle, the difference of the distances from each point along the extension of the base to the lateral sides is constant, namely it is congruent to the altitude dropped to a lateral ide.

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Exercise 188 by blakesterblakester, 17 Sep 2017 15:02

Exercise 187: In an isoceles triangle the sum of the distances from each point of the base to the lateral sides is constant, namely it is congruent to the altitude dropped to the lateral side.

Let the isoceles triangle be $\triangle ABC$ with arbitrary base point $D$, altitude from $\angle A$ to side $BC$ is $AF$, segment $DG$ is distance from $D$ to $AB$, and segment $DH$ is distance from $D$ to side $BC$. $AF$ and $DH$ are perpendicular to $BC$ as given.
$AF \parallel DH$, $\S 73(1)$ two lines intersected by a third and some corresponding angles are congruent are parallel lines.

Construct a line through point $D$ and parallel to $BC$, its intersection with $AB$ is point $J$ and intersection with $AF$ is point $K$. $KF = DH$, $\S 85$) parallel segments cut-out by parallel lines.
$\angle ADJ = \angle ACB$, $\S 77(1)$ parallels transversed corresponding angles are congruent. It follows that $\angle GAD = \angle ADJ$, $\S 35(2)$ isoceles base angles are congruent.
$\angle DKA = \angle CFK = d$, $\S 77(1)$. $\triangle AKD = \triangle DGA$, $\S 55(1)$ right triangles with a congruent hypotenuse and acute angle are congruent.

Thus, $AK = DG$, the altitude $AF = AK + KF$, and $DG + DH = AF$.

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Exercise 187 by blakesterblakester, 17 Sep 2017 15:02

Exercise 186: Through the vertices of a triangle, the lines parallel to the opposite sides are drawn. Prove that the triangle formed by these lines consists of four triangles congruent to the given one, and that each of its sides is twice the corresponding side of the given triangle.

Given $\triangle ABC$. Line through point $A$ parallel to $BC$ is $DE$. Line through point $B$ parallel to $AC$ is $DF$. Line through point $C$ and parallel to $AB$ is $EF$.

$\angle BAC = \angle DBA$ and $\angle DAB = \angle CBA$; $\S 77(2)$ alternate angles in parallels transversed are congruent.
$\triangle DBA = \triangle ABC$, $\S 40(2) SAS$. $DA = BC$. $AB = CF$, $\S 85$) parallel segments cut-out by parallel lines.
$BFAC$ is a parallelogram. $\triangle ABC = \triangle BFC$, $\S 40(3)$ SSS. $DB = AC = BF$; $DF = 2(AC)$

$AB = CE$, $\S 77(2)$. $BCAE$ is a parallelogram. $BC = AE$. $\triangle ABC = \triangle AEC$, $\S 40(3)$ SSS.
$AE = BC = AD$; $DE = 2(BC)$.
$CF = AB = CE$; $EF = 2(AB)$.

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Exercise 186 by blakesterblakester, 17 Sep 2017 15:01
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