Exercise 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.