Finding the length of one side of cyclic trapezium if given length of two adjacent side and the diagonal between them.

Question

I'm solving some practice problems to prepare for a competitive exam . Here is one which I'm trying to do for some time but still haven't found a solution to :

"$ABCD$ is a cyclic trapezium where $AB$ is parallel to $CD$ . If $AB = 4 , AD = 5 , BD = 7 $ find the length of $CD$ ."

Here is the drawing that I made :

I actually did the construction on paper and got that $CD$ is close to 6 , But I've no idea how to actually solve the problem. Are there any special results for cyclic trapeziums which I have to use here ? Please help.

Answer 1

Note that $BC=5$ and $AC=7$. Let our unknown side be $x$. Then by Ptolemy's Theorem, $$7\cdot 7=4x+5\cdot 5.$$

Remark: We could have used more basic machinery (similar triangles). However, Ptolemy's Theorem crops up in many problems about cyclic quadrilaterals, and can be considered essential knowledge.

One interesting thing about the theorem is that it was used by Ptolemy for applied purposes. He was primarily an astronomer, and used the theorem as a tool to draw up a "Table of Chords," a precursor of tables of the sine function.

Answer 2

Draw a point $E$ on $DC$ such that $BE$ is perpendicular to $DC$. Let $EC = x$. Since this is a cyclic trapezium, its also isosceles, and hence $BC=5$ and $DE=4+x$. Hence, $$\begin{aligned}BE^2=BD^2-DE^2&=BC^2-CE^2\\7^2-(4+x)^2&=5^2-x^2\\x&=1\end{aligned}$$ and hence you get $CD=(4+1)+1=6$.