Question

In the following figure, ∠BAD = ∠ACD. Prove that DA² = CD × BD.

Answer 1

Given: ∠BAD = ∠ACD

Since B, C, and D are collinear, the rays DB and DC lie on the same straight line.

Hence, ∠BDA = ∠ADC

So, in triangles △BAD and △ACD,

• ∠BAD = ∠ACD
• ∠BDA = ∠ADC

Therefore, △BAD ∼ △ACD by AA similarity.

Now let’s write the corresponding sides:

BD ↔ AD, AD ↔ CD

Hence, `(BD)/(AD) = (AD)/(CD)`

Cross-multiplying,

AD² = CD × BD

Thus,

DA² = CD × BD