Question

If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that `AQ = 1/2 (BC + CA + AB)`

Answer 1

BC + CA + AB

= (BP + PC) + (AR – CR) + (AQ – BQ)

= AQ + AR – BQ + BP + PC – CR

∵ From the same external point, the tangent segments drawn to a circle are equal.

From the point B, BQ = BP

From the point A, AQ = AR

From the point C, CP = CR

∴ Perimeter of ΔABC,

i.e., AB + BC + CA = 2AQ – BQ + BQ + CR – CR

⇒ 2AQ = AB + BC + CA

⇒ `AQ = 1/2 (BC + CA + AB)`

Hence proved.