Question

A circle is touching the side BC of a ∆ABC at P and is touching AB and AC when produced to Q and R, respectively.
Prove that $AQ=\frac{1}{2}$ (perimeter of ∆ABC).
Figure

Answer 1

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴
AQ = AR ..........(i) [tangent from A]
BP = BQ ...........(ii) [tangent from B]
CP = CR ...........(iii) [tangent from C]
Perimeter of ΔABC
= AB + BC + AC
= AB + BP + CP + AC
= AB + BQ + CR + AC [using (ii) and (iii)]
= AQ + AR
= 2AQ [ using (i)]
∴ $\mathrm{AQ}=\frac{1}{2}\left(\mathrm{Perimeter}\mathrm{of}△\mathrm{ABC}\right)$