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 $AQ = \frac{1}{2}(BC + CA + AB)$.

(N/A) As shown in the figure,the circle touches the sides of the triangle $ABC$ at points $P$,$Q$,and $R$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,we have:
$BQ = BP$
$CP = CR$
$AQ = AR$
Now,consider the perimeter of the triangle $ABC$:
$Perimeter = AB + BC + AC$
$Perimeter = AB + (BP + CP) + AC$
Substituting the values from the tangent theorem:
$Perimeter = AB + BQ + CR + AC$
$Perimeter = (AB + BQ) + (AC + CR)$
$Perimeter = AQ + AR$
Since $AQ = AR$,we can write:
$Perimeter = AQ + AQ = 2AQ$
Thus,$2AQ = AB + BC + AC$
$AQ = \frac{1}{2}(AB + BC + AC)$