Draw an isosceles $\mathrm{\Delta }\mathbf{ABC}$whose vertex angle, $\angle \mathbf{A}$has measure 80.

a. Draw $\overline{\mathrm{AX}}$the bisector of an exterior angle at $A$Is $\overline{\mathrm{AX}}\parallel \overline{\mathrm{BC}}$? Explain.

b. Would your answer change if the measure of$\angle \mathbf{A}$ changed?

From the figure it can be observed$\angle CAX=\angle XAY$ that by the definition of angle bisector.

By angle addition postulate,

$$\begin{gathered} m\angle BAC+m\angle CAX+m\angle XAY=180 \\ 80+m\angle CAX+m\angle CAX=180 \\ 2m\angle CAX=100 \\ m\angle CAX=50 \end{gathered}$$

From the figure it can be observed that, $\overline{AB}\cong \overline{AC}$, then by isosceles triangle theorem, $\angle B\cong \angle C$.

Consider $\mathrm{\Delta }ABC$then by angles sum theorem,

$$\begin{gathered} \angle BAC+\angle B+\angle C=180 \\ 80+\angle C+\angle C=180 \\ 2\angle C=100 \\ \angle C=50 \end{gathered}$$

As$m\angle CAX=50$ and$m\angle C=50$ implies that$\angle CAX=\angle C$ and they make a pair of alternate interior angles. Alternate interior angles formed when two parallel lines are cut by transversal, therefore, $\overline{AX}\parallel \overline{BC}$.

Suppose the measure of$\angle A$ is changed to 60. From the figure it can be observed that$\angle CAX=\angle XAY$ by the definition of angle bisector.

By angle addition postulate,

$$\begin{gathered} m\angle BAC+m\angle CAX+m\angle XAY=180 \\ 60+m\angle CAX+m\angle CAX=180 \\ 2m\angle CAX=120 \\ m\angle CAX=60 \end{gathered}$$

From the figure it can be observed that, $\overline{AB}\cong \overline{AC}$, then by isosceles triangle theorem, width="75">∠B≅∠C.

Consider$\mathrm{\Delta }ABC$ then by angles sum theorem,

$$\begin{gathered} \angle BAC+\angle B+\angle C=180 \\ 60+\angle C+\angle C=180 \\ 2\angle C=120 \\ \angle C=60 \end{gathered}$$

As$m\angle CAX=60$ and$m\angle C=60$ implies that$\angle CAX=\angle C$ and they make a pair of alternate interior angles. Alternate interior angles formed when two parallel lines are cut by transversal, therefore, $\overline{AX}\parallel \overline{BC}$. Therefore, even if the measure of$\angle A$ is changed, the answer will remain the same.