Question

Quad. $ABCD$ is circumscribed about a circle. Discover and prove a relationship between $AB+DC$ and $AD+BC.$

Short answer

$AB+DC$and $AD+BC$

Step-by-step solution

Step 1. Given information.

Let $ABCD$ be a quadrilateral circumscribing the circle with centre $O$. The quadrilateral touches the circle at point $P,Q,R$ and $S$.

Step 2. Formula used.

Lengths of tangents drawn from external point are equal.

Step 3. Proof.

Consider the figure below,

According to theorem, lengths of tangents drawn from external point are equal.

Then,

$\begin{array}{l}AP=AS\\ BP=BQ\\ CR=CQ\\ DR=DS\end{array}$

Step 4. Adding above equations.

$\begin{array}{l}AP+BP+CR+DR=AS+BQ+CQ+DS\\ \left(AP+BP\right)+\left(CR+DR\right)=\left(AS+SD\right)+CQ+BQ\\ AB+CD=AD+BC\end{array}$

Hence, it is proved that, $AB+CD=AD+BC$