Question

In the given Δ ABC, AB = AC. The angle bisector of ∠A meets the side BC at the point E.
The measure of ∠BAE$ is degrees.

Answer 1

The given triangle is as follows:

$\text{Given:}$ $AB=AC$

We know that angles opposite to equal sides are equal in measure.
$\therefore$ $m\angle B=m\angle C={50}^o$

Now, sum of all three interior angles of a triangle is ${180}^o$
$\therefore$ $m\angle A+m\angle B+m\angle C={180}^o$
$\Rightarrow$ $m\angle A+{50}^o+{50}^o={180}^o$
$\Rightarrow$ $m\angle A={80}^o$

Now, the line $AE$ divides the $\angle A$ in two equal halves, $\angle BAE\text{and}\angle CAE$.

Hence, $m\angle BAE=\frac{m\angle A}{2}=\frac{{80}^o}{2}={40}^o$