Question

In the given figure, ABC is a triangle in which AB = AC and D is a point on AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.

Answer 1

In the given figure, $DE\left|\right|BC$.
$$\begin{gathered} \angle ADE=\angle ABC \\ \angle AED=\angle ACB \end{gathered}$$ (Corresponding angles) ...(i)

But in triangle ABC, AB = AC.
$\therefore \angle ABC=\angle ACB$ ...(ii)

From (i) and (ii), we have:

$$\begin{gathered} \angle ADE=\angle ABC \\ \angle AED=\angle ABC \end{gathered}$$
$\Rightarrow \angle ADE=\angle AED$

Therefore, triangle ADE is also isosceles.
Or AD = AE
Hence, proved.