In Figure 6, a quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = BC + AD$.

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Question

In Figure 6, a quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = BC + AD$.

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Accepted Answer

Step 1: Let the circle touch the sides $AB$, $BC$, $CD$ and $AD$ at $P$, $Q$, $R$ and $S$ respectively.
Step 2: Since tangents from an external point are equal: $AP = AS$, $BP = BQ$, $DR = DS$, $CR = CQ$. Adding: $(AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ)$
Step 3: $	herefore AB + CD = BC + AD$

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