Question

If a circle is touching the side BC of △ ABC at P and is touching AB and AC produced at Q and R respectively (see the figure).

Prove that AQ = \(\frac12\)​ (perimeter of △ ABC).

Answer 1

Given: A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.

Proof: Lengths of tangents drawn from an external point to a circle are equal.

⇒ AQ = AR, BQ = BP, CP = CR.

Perimeter of △ABC = AB + BC + CA

= AB + (BP + PC) + (AR – CR)

= (AB + BQ) + (PC) + (AQ – PC) since [AQ = AR, BQ = BP, CP = CR]

= AQ + AQ

= 2AQ

⇒ AQ = \(\frac12\)​(Perimeter of △ABC)

∴ AQ is the half of the perimeter of △ABC