91Ó°ÊÓ

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

By angle addition postulate,

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

From the figure it can be observed that, $\stackrel{¯}{AB}â‰\dots \stackrel{¯}{AC}$, then by isosceles triangle theorem, $∠Bâ‰\dots ∠C$.

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

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

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

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

By angle addition postulate,

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

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

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

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

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