Methods Of Construction And Symmetries

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Methods Of Construction And Symmetries

page revision: 0, last edited: 15 Sep 2008 08:58

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## Have a question?

The unmarked bold line is referred to as "CD" in the text.

ReplyOptionsIn my 2006? printing, figure 108 point $C'$ seems to be labeled $C$ (sans prime).

ReplyOptionsExercise 207:

Construct a triangle, given: (a) its base, the altitude, and a lateral side; (b) its base, the altitude, and an angle at the base; (c) an angle, and two altitudes dropped to the sides of this angle; (d) a side, the sum of the other two sides, and the altitude dropped to tone of these sides; (e) an angle at the base, the altitude, and the perimeter.(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.

ReplyOptionsExercise 208:

Construct a quadrilateral given three of its sides and both diagonals.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, $C'D' = CD$, $C'Y = AD$, $YZ = BC$, $D'Z = AB$, $C'Z = BD$, $D'Y = AC$.

ReplyOptionsExercise 209:

Construct a parallelogram given: (a) two non-congruent sides and a diagonal; (b) one side and both diagonals; (c) the diagonals and the angle between them; (d) a side, the altitude, and a diagonal (is this always possible?).(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.

ReplyOptionsExercise 210:

Construct a rectangle, given a diagonal and the angle between the diagonals.The unknown diagonal is congruent to the given diagonal, $\S 90$. This is the same problem as #209(c).

ReplyOptionsExercise 211:

Construct a rhombus given: (a) its side and a diagonal; (b) both diagonals; (c) the distance between two parallel sides, and a diagonal; (d) an angle, and the diagonal passing through its vertex; (e) a diagonal, and an angle opposite it; (f) a diagonal, and the angle it forms with one of the sides.(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.

ReplyOptionsExercise 212:

Construct a square, given its diagonal.$\S 92$) 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.

ReplyOptionsExercise 213:

Construct a trapezoid given: (a) its base, an angle adjacent to it, and both lateral sides (there can be two solutions, one or none); (b) the difference between the bases, a diagonal, and lateral sides; (c) the four sides (is this always possible?); (d) a base, its distance from the other base, and both diagonals (when is this possible?); (e) both bases and both diagonals (when is this possible?).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.

ReplyOptionsExercise 214:

Construct a square, given: (a) the sum of a diagonal and a side; (b) the difference of a diagonal and an altitude.(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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