Question

On the sides, AB and AC of triangle ABC, equilateral triangles ABD and ACE are drawn. Prove that: $\angle \text{CAD}=\angle \text{BAE}$

Answer 1

Given:$△ABD$ and $△ACE$ are equilateral triangle.

$\therefore$ In $△ABD$

$\angle BAD=\angle ABD=\angle BDA={60}^{\circ }$ .....$\left(1\right)$

In $△ACE$

$\angle AEC=\angle ACE=\angle ACE={60}^{\circ }$ .....$\left(2\right)$

To prove:$\angle CAD=\angle BAE$

Proof:$\angle CAD=\angle BAC+\angle DAB$

$\Rightarrow \angle CAD=\angle BAC+{60}^{\circ }$ using $\left(1\right)$

$\Rightarrow \angle CAD=\angle BAC+\angle CAE$ using $\left(2\right)$

$\Rightarrow \angle CAD=\angle BAE$

Hence proved.