The term 2db should be 2dn.

Exercise 153: *Compute the angle between two medians of an equilateral triangle.*

Equilateral triangle is isoceles by ($\S 33$) definition having two congruent sides. $\S 35(1)$ - Median, altitude, and bisector of vertex in isoceles triangle are one in the same.

Let the point of intersection of the medians be $D$, the median point along side $AC$ be $F$ . The median from $B$ to $AC$ is perpendicular, $\angle AFD = d$ . Median from $A$ to side $BC$ bisects $\angle A$ , which has measure of 60 degrees. Therefore $\angle DAF = 30^{\circ}$.

$\S 81$) Angle sum of a triangle is $2d = 180^{\circ}$. $\triangle FAD$ has a 90 degree and a 30 degree angle, so the remaining angle, $\angle ADF$, must be 60 degrees. The other angle between the medians, $\angle FDG$, is supplementary ($\S 22$) to $\angle ADF$ and must be measure 120 degrees.

Exercise 154: *Compute the angle between bisectors of actute angles in a right triangle.*

Let the right triangle be $\triangle ABC$. $\S 81$) sum of angles in a triangle is $2d$.

Acute $\angle A$ of the triangle is bisected into two congruent summands, $\alpha$. The other acute angle, $\angle B$, bisected into two congruent summands, $\beta$. These bisectors intersect at point $D$.

In $\triangle ADC$, $\alpha + \beta + \delta = 2d$.

In $\triangle ABC$, $2\alpha + 2\beta = d$.

$\alpha + \beta = \frac{d}{2}$

Thus, $\frac{d}{2} + \delta = 2d = \frac{4d}{2}$.

$\delta = \frac{3d}{2}$

Supplementary angle ($\S 22$) $\gamma = \frac{d}{2}$

Exercise 155: *Given an angle of an isoceles triangle, compute the other two. Consider two cases: the given angle is (a) at the vertex, or (b) at the base.*

(a)

Let the given vertex angle be $\alpha$ and a base angle be $\beta$.

$\alpha + 2\beta = 2d$ : $\S 81$) Triangle sum is $2d$ .

$\frac{\alpha}{2} + \beta = d$

$\beta = d - \frac{\alpha}{2}$ is the measure of both base angles.

(b)

Let the given base angle be $\angle \beta$ and the vertex angle be $\angle \alpha$.

$\alpha + 2\beta = d$

$\alpha = 2d - 2\beta$ is the measure of the vertex.

Exercise 156: *Compute interior and exterior angles of an equiangular pentagon.*

$\S 82$) Angle sum of a polygon = $2d(n - 2)$.

Given all interior angles are congruent in an equiangular figure, any interior angle has the measure: $\frac{2d(n - 2)}{n}$ .

The angle exterior to any of these is a supplementary angle ($\S 22$): $2d - (2d - \frac{4d}{n}) = \frac{4d}{n}$

Exercise 157: *Compute angles of a triangle which is divided by one of its bisectors into two isoceles triangles. Find all solutions.*

(a) Case where the bisected angle part is congruent to the other angle opposite the bisector side in the created sub isoceles triangle.

Bisector = $BD$. $\angle ABD = \angle CBD$.

$\angle ABD = \angle DAB$ in $\triangle ABD$

$\angle CBD = \angle DCB$ in $\triangle CBD$

It follows that $\angle DAB = \angle DCB$ and the original $\triangle ABC$ is isoceles, $AB = CB$ ($\S 33$ - isoceles triangle has two congruent sides).

$AD = BD = CD$

$\S 40(3)$ SSS; $\triangle ABD = \triangle CBD$

$\angle ADB = \angle CDB$. $\S 22$) supplementary angles sum to $2d$, so each of these angles measures $d$.

But $\angle DAB = \angle ABD$ and $\S 81$) triangle angle sum is $2d$: $\angle DAB + \angle ABD = d$ .

$\angle DAB = \angle ABD = \frac{d}{2}$.

Angles in $\triangle ABC$ are $\angle CAB = \frac{d}{2}$, $\angle ACB = \frac{d}{2}$, and $\angle ABC = d$.

(b) Case where the bisected angle part is NOT congruent to the other two angles in the created sub isoceles triangle.

Bisector = $BD$. $\angle ABD = \angle CBD$.

The two isoceles triangles created by the bisector are: $\triangle ABD$, $\angle BAD = \angle BDA$ ; and $\triangle CBD$, $\angle BDC = \angle BCD$.

$AB = DB$ and $DB = CB$ as these are the sides in the created isoceles triangles that are opposite the congruent angles. It follows that $AB = CB$.

The original $\triangle ABC$ is isoceles. $\angle BAD = \angle BCD$.

$\S 40(3)$ ASA, $\triangle ABD = \triangle CBD$.

$\angle ADB = \angle CDB$ and $\S 22$) supplementary angles sum to $2d$.

But $\angle BAD = \angle ADB$, so $\angle BAD + \angle ADB = 2d$. This is impossible as the angle sum of $\triangle ABD$ is $2d$ ($\S 81$) and that must include the measure of $\angle ABD$ .

(c) Case where the bisected angle part is congruent to the angle in the sub isoceles triangle at the other end of the bisector.

Bisector = $BD$. $\angle ABD = \angle CBD = \angle CDB = \angle ADB$.

$\S 22$) supplementary angles are congruent to $2d$. $\angle ADB$ and $\angle CDB$ are supplementary and mutually congruent. $\angle ADB = \angle CDB = d$.

But in $\triangle ABD$, $\angle ADB = \angle ABD$, so $\angle ABD = d$.

$\S 81$) Triangle angle sum is $2d$. $\angle ABD + \angle ADB = 2d$ contradicts $\angle ABD + \angle ADB + \angle BAD = 2d$.

Exercise 158: *Prove that if two angles and the side opposite to the first of them in one triangle are congruent respectively to two angles and the side opposite to the first of them in another triangle, then such triangles are congruent.*

Let the triangles be $\triangle ABC$ and $\triangle A'B'C'$ where $\angle B = \angle B'$, $\angle A = \angle A'$, $\angle AC = \angle A'C'$.

$\S 81$ corollary #2) If two angles in one triangle are respectively congruent to two angles in another triangle, then the remaining angles are congruent. $\angle C = \angle C'$.

$\S 40(2)$ ASA : $\triangle ABC = \triangle A'B'C'$.

Exercise 159: *Prove that if a leg and the acute angle opposite to it in one right triangle are congruent respectively to a leg and the acute angle opposite to it in another right triangle, then such triangles are congruent.*

Same as #158, AAS.

Exercise 160: *Prove that in a convex polygon, one of the angles between the bisectors of two consecutive angles is congruent to the semisum of these two angles.*

Let the polygon be $\triangle ABC$. $\angle A$ bisected to form two congruent sub-angles $\alpha$. $\angle B$ bisected to form two congruent sub-angles $\beta$.

$\angle \epsilon$ is exterior to $\angle BAC$. $\angle \eta$ is exterior to $\angle BCA$.

Opposite to $AC$ is $\angle \alpha\gamma\beta$ at the intersection of the bisectors. Supplementary to this angle is $\theta$.

$\angle \beta + \angle \gamma = \angle \epsilon + \angle \alpha$; $\S 81$ angle sum of a triangle.

$\angle \alpha + \angle \gamma = \angle \beta + \angle \eta$

$2 \angle \gamma + \angle \beta + \angle \alpha = \angle \epsilon + \angle \eta + \angle \beta + \angle \alpha$

$2 \angle \gamma = \angle \epsilon + \angle \eta$

$\angle \gamma = \frac{\angle \epsilon + \angle \eta}{2}$

$\angle \epsilon + 2 \angle \alpha = 2d$

$\angle \epsilon = 2d - 2 \angle \alpha$

$\angle \eta + 2 \angle \beta = 2d$

$\angle \eta = 2d - 2 \angle \beta$

$\angle \gamma = \frac{2d - 2 \angle \alpha + 2d - 2 \angle \beta}{2}$

$\angle \gamma = d - \angle \alpha + d - \angle \beta$

$\angle \gamma = 2d - \angle \alpha - \angle \beta$

$\angle \gamma + \angle \theta = 2d$

$\angle \gamma + \angle \theta = 2d$

$2d - \angle \alpha - \angle \beta + \angle \theta = 2d$

$\angle \theta = \angle \alpha + \angle \beta$; the latter is the semi-sum of $\angle A$ and $\angle B$.

Exercise 161: *Given two angles of a triangle, construct the third one.*

Given $\angle A$ and $\angle B$. $\S 81$) The angle sum of a triangle is $2d$. Use a straight edge to construct the straight angle $CD$ from one side of $\angle A$. Place $\angle B$ with the vertex at the vertex of $\angle A$, one side along and in the same direction as the side $\angle A$ not along $CD$, and its other side external to $\angle A$. The third angle is composed of this side of $\angle B$ in the exterior of $\angle A$ and the side along $CD$ in the direction away from $\angle A$.

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