On page 50, line 9, $DE=EF$ should read $DF=EF$

Exercise 115: *The sum of two, three, or more given angles.*

Solution assumes that neither angle being summed is a full angle. In the case that any of the given angles are full, the construction is a full angle.

$\S 15$) Sum of angles. Choose two of the angles ($\angle AOB$ and $\angle COD$) and construct a circle around their vertices using the same arbitrary radius. In the circle about $\angle COD$ draw chord $EF$ between the points where side $OC$ and side $OD$ intersect the circle, respectively. Note if chord $EF$ is within the interior or exterior of $\angle EOF$.

Set the compass step congruent to $EF$. Place the pin of the compass at point $B$. If chord $EF$ was in the interior, sweep the free leg of the compass in the direction of the circle leading away from side $OB$ and into the exterior of $\angle AOB$. Otherwise sweep in the direction of the circle leading into the interior of $\angle AOB$. The point of intersection of the compass mark and the circle is point $D'$. If $D'$ is within the interior of $\angle AOB$, the sum is greater than the full angle and cannot be represented in a single figure. Otherwise draw radius $OD'$.

$\angle BOD' = \angle COD$. $\angle AOD' = \angle AOB + \angle BOD'$.

For 3 or more given angles, sum two of the angles then add this sum to the third. Repeat using the previous sum for the fourth and so on.

Exercise 116: *The difference of two angles.*

Construct a circle of the same arbitrary radius about $\angle AOB$ and $\angle COD$. In the circle containing $\angle COD$ draw chord $DE$ between the circle intersection points of side $\overrightarrow{OC}$ and $\overrightarrow{OD}$, points $D$ and $E$ respectively. Set the compass step congruent to $DE$. Note whether chord $DE$ is interior or exterior to $\angle COD$.

If $\angle AOB \leq$ the straight angle AND $\angle COD \leq$ the straight angle, place the compass pin at the point where $\overrightarrow{OB}$ intersects the circle, point $B'$. Sweep the marking leg in the direction of the interior of $\angle AOB$ and mark on the circle a point $Z$. $\angle ZOA$ is the difference of the angles. If $\angle COD > \angle AOB$, this is negative.

If $\angle AOB \leq$ the straight angle XXX

If the two angles are congruent the difference is no angle, stop.

Let the greater of the two angles be $\angle AOB$ and the lesser be $\angle COD$. Construct a circle of the same arbitrary radius about both angles. $\angle AOB$ intersects the circle at points $A'$ and $B'$. $\angle COD$ at points $C'$ and $D'$.

In the circle containing $\angle COD$ construct chord $C'D'$ and set the compass step congruent to this chord. Note when $\angle COD$ is greater than the straight angle this chord is through the exterior angle.

If $\angle COD \leq$ the straight angle, place the compass pin at point $B'$ and sweep the free leg in the direction of the interior of $\angle AOB$. Mark the compass intersection with the circle point $Z$. $\angle AOZ$ a sub-sector of the interior of $\angle AOB$ is the difference.

If $\angle COD >$ the straight angle, place the compass pin at point $B'$ and sweep the free leg in the direction of the exterior of $\angle AOB$. Mark the compass intersection with the circle point $X$. $\angle AOX$ a sub-sector of the interior of $\angle AOB$ is the difference.

If $\angle COD$ is equal to some number of whole angles, the figure likely does not change as $\angle AOB$ is greater than the whole angle and this probably isn't represented. If there is an indication of whole angles, adjust it minus the number of whole angles.

If $\angle COD$ is greater than some number of whole angles, first apply the "equal to some number of whole angles" rule to eliminate the whole angle summands then the pertinent less than whole angle rules for the remaining component.

Exercise 117: *Two angles whose sum and difference are given.*

Given: $\angle A + \angle B$ $\angle A - \angle B$

If the given difference is zero / no angle at all, the two angles are congruent. Bisect the given sum, the two newly formed angles are $\angle A$ and $\angle B$, respectively.

Given the sum $\angle A + \angle B$ and difference $\angle A - \angle B$. The sum of these given composite angles is: $(\angle A + \angle B) + (\angle A - \angle B) = (2)\angle A$.

Construct a circle of the same arbitrary radius about both given angles. Set the compass step congruent to the chord between the points of intersection of the given difference angle and its circle. Place the compass pin at one of the points of intersection of the given sum angle and its circle. If the given difference angle is greater than the straight angle, sweep the compass marking leg in the direction of the interior of the given sum angle. Otherwise sweep in the direction of the exterior. The sum of the given composite is $\angle C$, which has sides where the compass pin was set and where the marking leg intersected the circle, and interior starting at the compass pin side and continuing in the direction of the interior of the original sum angle. $\angle C = (2)\angle A$. Bisect $\angle C$ to obtain $\angle A$. If $\angle C$ is greater than the whole angle, this must be accounted.

Construct a duplicate of the given sum angle with a circle about it using the same arbitrary radius first used. Let the points of intersection with the circle be $Y$ and $Z$, and vertex $O$. Place the compass pin at point $Y$ and sweep in the direction of the interior of the given sum angle. The marking leg intersection with the circle is point $X$. $\angle YOX = \angle A$, $\angle ZOX = \angle B$.

Exercise 118: *Divide an angle into 4, 8, 16 congruent parts.*

Let the given angle be $\angle O$. Using the angle bisection method described in $\S 64$, which renders $\angle A$ and $\angle Z$. By the halving definition of a bisector, $\S 15$, these new half angles are congruent. Repeat the same bisection technique on these angles to produce angles $\angle B$ and $\angle C$, and $\angle X$ and $\angle Y$, respectively. These new angles are congruent halves of congruent halves and must be congruent as well. This satisfies the 4 congruent parts.

Continuing to bisect the most recent step's congruent half angles doubles the congruent parts. Thus, continue the process twice more to produce 8 then 16 parts.

Exercise 119: *A line in the exterior of an angle passing through its vertex and such that it would form congruent angles with the sides of this angle.*

It is only possible for the line to lie entirely in the exterior of the given angle if this angle less than a right angle. Otherwise part of the line will exist as a ray within the interior of the angle.

In this case the exterior line is constructed by first using the bisection method of $\S 64$ to bisect the given angle. At the vertex point on this bisector erect a perpendicular using the method in $\S 65$. The erected perpendicular is the line that satisfies the question. The angles between the erected perpendicular and the bisector are right, $\S 23$. The halves of the given angle formed by the bisector are congruent, $\S 15$. Thus, the remaining angle in each right angle previously described, with one side shared with a bisect half and the other along the erected perpendicular, is congruent to one another.

If it is only necessary that part of the line satisfy the question:

Bisect the given angle using the method in $\S 64$. Continue the bisector ray into the exterior of the given angle. I.e. bisect the exterior angle. This ray / bisector in the exterior satisfies the question, $\S 23$.

Exercise 120: *A triangle: (a) given two sides and the angle between them; (b) given one side and both angles adjacent to it; (c) given two sides and the angle opposite to the greater one of them; (d) given two sides and the angle opposite to the smaller one of them (in this case there can be two solutions, or one, or none).*

(a)

Use a straight edge to connect the terminal points of the given sides opposite the end with the given angle. $\S 4$) straight line through two points, $\S 32$) three sided polygon is a triangle.

(b)

Uncerain if the given angles are partially constructed on the ends of the given side. Assuming they are, continue the not given sides of the angles until they intersect on one side of the given triangle side. $\S 4$) straight line axiom; two disctinct lines can intersect at most at one point. $\S 32$) three-sided polygon is a triangle.

(c)

The given angle will be at one end of the lesser given side, necessary to be opposite the greater side. Set the compass step congruent to the greater given side. Place the pin at the end of the shorter given side, opposite the given angle, and mark a circle. Using a straight edge continue the not given side of the given angle through the previously constructed circle. The point of intersection is the unknown axis of the triangle, $\S 4$) straight line axiom.

Using a straight edge connect the point where the compass pin was located to this intersection point. This is the given greater side.

(d)

The given angle must be at one end of the given greater side in order to be opposite the smaller given side. Set the compass step congruent to the smaller given side. Place the pin at the end of the greater given side, opposite end end with the given angle, and mark a circle. Extend the not given side of the given angle using a straight edge.

No solution case: The described circle and the not given side do not intersect, so no triangle is formed.

One solution case: The not given side intersects the described circle at only one point (tangent line). This intersection is the unknown axis of the triangle. Using a straight edge connect the point where the compass pin was placed to this intersection point.

Two solution case: The not given side intersects the described circle at two points. Either of these points of intersection can be the unknown axis of the triangle. Use a straight edge to connect the point where the compass pin was placed to the desired third axis point.

Exercise 121: *An isoceles triangle: (a) given its base and another side; (b) given a base and a base angle; (c) given its base angle and the opposite side.*

$\S 33$) isoceles triangle has two congruent sides.

(a)

Thus the not given side is congruent to the given non-base side. Unless the base and the given side are congruent, in which case there are infinitely many triangles that could be constructed.

Set the compass step congruent to the given non-base side. At both ends of the given base mark a semicircle on the same side of the base. The point of intersection of these two arcs is the unknown vertex o the triangle. Using a straight edge connect the end points of the base to this intersection point.

If the given non-base side is less than half of the base, there is no solution; $\S 48$.

(b)

$\S 35(2)$ - Angles at base of isoceles triangle are congruent.

Mark an angle congruent to the given angle at the end of the base opposite the given angle and on the same side of the base. Using a straight edge extend the sides of the base angles not along the base until they intersect. $\S 4$) straight line axiom. This point of intersection is the unknown vertex of the triangle.

The given angle must be less than right. Otherwise the non-base sides never intersect or won't intersect on the expected side of the base, and thus there would be no solution.

(c)

Using a straight edge and compass extend the non-base side of the given base angle to be congruent to the given side. Set the compass step congruent to the given side. Place the pin at the end of the previously extended side of the given base angle, opposite the base angle. Mark the circle about this point. Using a straight edge extend the base side of the given base angle until it interrsect the marked circle. This point of intersection is the other end of the base. Using a straight edge connect this point of intersection to where the compass pin was placed to mark the circle. The triangle has been constructed.

Exercise 122: *A right triangle: (a) given both of its legs; (b) given one of the legs and the hypotenuse; (c) given one of the legs and the adjacent acute angle.*

$\S 33$) In a right triangle the sides of the right angle are called the legs.

(a)

Using a straight edge connect the end points of the given legs opposite the ends with the right angle. $\S 4$) straight line axiom. The right triangle has been constructed.

(b)

Using a straight edge and compass, add a segment congruent to the given leg, at one end of the given leg, point $R$, and along the same line. Use the method in $\S 67$ to draw the perpendicular through the midpoint, $R$, of the resultant segment. Set the compass step congruent to the given hypotenuse. Place the compass pin at the end point of the given leg opposite point $R$ (where the right angle is). Mark a semicircle on one side of the given leg. If necessary extend the previously drawn perpendicular to intersect the semicircle. Use a straight edge to connect the point where the compass pin was placed to draw the semicircle and this point of intersection. The triangle has been constructed and the not given leg is the segment from point $R$ to the intersection.

(c)

Using a compass and straight edge, add a segment congruent to the given leg to the end of the the given leg opposite the given angle and along the same line. Use the method in $\S 67$ to draw the perpendicular to the midpoint of the resultant segment. This midpoint is $R$, the end of the given leg. Using a straight edge extend the not given side of the given angle and the drawn perpendicular until they intersect. $\S 4$) straight line axiom. The triangle has been constructed.