Quadrilateral $ABCD$ is circumscribed about circle $O$. If $AB=52$, $BC = 40$, and $AD =48$. How can I find the length of $CD$? I know the answer should be $36$, but I am unsure of how to arrive at that answer.
2 Answers
$\begingroup$Hint: For each vertex, the two segments connecting that vertex to a point of tangency in the circle are the same length. (Why?)
So, for instance, if the points of tangency are $P \in \overline{AB}$, $Q \in \overline{BC}$, $R \in \overline{CD}$ and $S \in \overline{AD}$, then $\overline{AP}$ and $\overline{AS}$ have the same length.
$\endgroup$ $\begingroup$Label the tangent points as shown:
Clearly $AE=AH, BE=BF, CF=CG, DG=DH$
Then
$\begin{align}
CD &=CG+DG \\ &= CF+DH\\ &= (BC-BF)+(AD-AH)\\ &= BC + AD - (BF+AH)\\ &= BC + AD - (BE+AE)\\ &= BC + AD - AB\
\end{align}$
(Perhaps more compactly, $AD+BC = AB+CD$ is almost immediately obvious)
$\endgroup$
