Question

Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that : `(AB)/(PQ) = (AD)/(PM)`.

Answer 1

Given, ΔABC ∼ ΔPQR

AD and PM are the angle bisectors of ∠A and ∠P respectively.

To prove: `(AB)/(PQ) = (AD)/(PM)`

Proof: Since, ΔABC ∼ ΔPQR

∴ ∠A = ∠P and ∠B = ∠Q

`(AB)/(PQ) = (BC)/(QR)`

∵ ∠A = ∠P

`\implies 1/2 ∠A = 1/2 ∠P`

`\implies` ∠BAD = ∠QPM

Now in ΔABD and ΔPQM

∠B = ∠Q

∠BAD = ∠QPM   ...(Proved)

∴ ΔBAD ∼ ΔQPM   ...(AA axiom)

∴ `(AB)/(PQ) = (AD)/(PM)`