Question

A circle touches the side $BC$ of a triangle $ABC$ at $P$ and the extended sides $AB$ and $AC$ at $Q$ and $R$ respectively. Prove that $AQ$ is equal to half the perimeter of $△ABC$

Answer 1

Lengths of $\Delta$ drawn from an external pt to a circle are equal.

$\Rightarrow$ AQ=AR, BQ=BP, CP=CR$

Perimeter of $\Delta ABC=AB+BC+CA$

$=AB+(BP+PC)+(AR-CR)$

$=(AB+BQ)+PC+(AQ-PC)$

$=AQ+AQ=2AQ$

$AQ=\frac{1}{2}$(perimeter of $\Delta ABC$).