Question

In the figure below, ΔABC is an isosceles triangle with AC=AB. D
and E is pointing on AB and AC respectively such that DE ∥ BC.
Prove that ΔADE is isosceles.

Answer 1

If ΔABC is an isosceles triangle and D & E are points on AB and AC respectively such that DE ∥ BC, then Δ ADE is isosceles.

Step-by-step explanation:

It is given that,

ΔABC is an isosceles triangle with

AB = AC …… (i)

DE // BC

Considering ΔADE & ΔABC, we have

∠A = ∠A ….. [common angle]

∠ADE = ∠ABC …… [since DE // BC so, both the angles are corresponding angles]

∴ By AA similarity, Δ ADE ~ Δ ABC

Since the corresponding sides of two similar triangles are proportional to each other.

∴ $\frac{AD}{AB}$ = $\frac{AE}{AC}$

⇒ AD = AE ……. [∵ AB = AC, from (i)]

∴ Δ ADE is an isosceles triangle  

Hence proved

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