86
In the given figure, $\triangle ABE \cong \triangle ACD$. Prove that $\triangle ADE \sim \triangle ABC$.
_q_1.png)
Show SolutionHide Solution↓
Given $\triangle ABE \cong \triangle ACD$
$\therefore AE = AD$ or $AD = AE$ ---- (1) ($1/2$)
and $AB = AC$ ---- (2) ($1/2$)
Dividing (1) by (2), we have
$\frac{AD}{AB} = \frac{AE}{AC}$ ($1/2$)
and $\angle DAE = \angle BAC$
$\therefore \triangle ADE \sim \triangle ABC$ ($1/2$)
$\therefore AE = AD$ or $AD = AE$ ---- (1) ($1/2$)
and $AB = AC$ ---- (2) ($1/2$)
Dividing (1) by (2), we have
$\frac{AD}{AB} = \frac{AE}{AC}$ ($1/2$)
and $\angle DAE = \angle BAC$
$\therefore \triangle ADE \sim \triangle ABC$ ($1/2$)