Question

In the following figure, DE || BC and DC || FE. Prove that AD² = AB × AF.

Answer 1

Given:

• DE || BC
• DC || FE

We have to prove:

AD² = AB × AF

⇒ In △ABC, since DE || BC,

△ADE ∼ △ABC

Therefore, `(AD)/(AB) = (AE)/(AC)`    ....(1)

⇒ In △ADC, since FE || DC,

△AFE ∼ △ADC

Therefore, `(AF)/(AD) = (AE)/(AC)`    ....(2)

Now, let’s compare (1) and (2),

`(AD)/(AB) = (AF)/(AD)`

Cross-multiplying,

AD² = AB × AF

Hence proved,