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Sum of Pair of Opposite Angles in Quadrilateral

In the given ...

Question

In the given figure, a circle touches the side $P Q$ of a $△ A P Q$ at $R$ and side $A P$ and $A Q$ produced at the point $B$ and $C$ . show that $A B = \frac{1}{2}$ (Perimeter of $△ A P Q)$

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Solution

$B P = B Q$
$C P = C R$
$A Q = A R$
$A B + B Q = A C + C R$
$A B + B P = A C + C P$
$△ A B C = A B + B C + C A$
$A B + (B P + P C) + C A$ \\
$(A B + B P) (P C + C A)$
$2 (A B + B P) (A B + B P = A C + C P)$
$2 (A B + B Q) = 2 A Q$
$P Q = \frac{P E R I M E T E R O F △ A B C}{2}$

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