Question

In the adjoining figure, `(AD)/(BD) = (AE)/(EC)` and ∠BDE = ∠CED, prove that ΔABC is an isosceles triangle.

Answer 1

Given, `(AD)/(BD) = (AE)/(EC)`   ...(1)

∠BDE = ∠CED   ...(∠1 = ∠2)

Now ∠1 + ∠3 = ∠2 + ∠4    ....(180° linear pair)

∴ ∠3 = ∠4    ...(2)

Now, from equation (1),

`(AD)/(BD) = (AE)/(EC)`

∴ By the converse of BPT

DE || BC

⇒ ∠3 = ∠5    ...[Corresponding angle]

∠4 = ∠6

∠5 = ∠6  ....[using (2)]

⇒ AC = AB    ....[Side opposite to equal angle]

∴ ΔABC is an isosceles triangle.