Question

In the following figure, AB = AC and AB² = BD × CE. Prove that ΔADB ~ ΔЕАС.

Answer 1

Given:

AB = AC and AB² = BD × CE, with D, B, C, E collinear.

Since AB = AC, in △ABC, base angles opposite equal sides are equal.

∴ ∠ABC = ∠BCA

Now D, B, C, and E are collinear, so BD is the extension of BC, and CE is the extension of CB.

Hence,

∠ABD = 180° − ∠ABC and ∠ACE = 180° − ∠BCA

But ∠ABC = ∠BCA. Therefore,

∠ABD = ∠ACE

Also, given

AB² = BD × CE

So,

AB × AB = BD × CE

Since AB = AC, we get

AB × AC = BD × CE

Dividing both sides by AC × CE, we obtain

`(BD)/(AC) = (AB)/(CE)`

Thus, in △ADB and △EAC,

`(BD)/(AC) = (AB)/(CE)` and ∠ABD = ∠ACE

Therefore, by SAS similarity,

△ADB ∼ △EAC

Hence proved.