Question
In the given figure $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and a circle passing through B andC intersects AB and AC at D and E respectively. Prove that DE || BC.
✓
Answer
$\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and circle passing through B and C intersects AB and AC at D and E.Since AB = AC
$\therefore\ \angle\text{ACB}=\angle\text{ABC}$
So, exterior
$\ \angle\text{ADE}=\angle\text{ACB}=\angle\text{ABC}$
$\therefore\ \angle\text{ADE}=\angle\text{ABC}$
$\Rightarrow\ \text{DE }||\text{ BC}$
$\therefore\ \angle\text{ACB}=\angle\text{ABC}$
So, exterior
$\ \angle\text{ADE}=\angle\text{ACB}=\angle\text{ABC}$
$\therefore\ \angle\text{ADE}=\angle\text{ABC}$
$\Rightarrow\ \text{DE }||\text{ BC}$
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