A Problem from BDMO $2008$ National.

$ABCD$ is a cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at $E$. $AB=39$; $AE=45$; $AD=60$; $BC=56$. Find the length of $CD$.

Since $ABCD$ is a cyclic quadrilateral So, Ptolemy's theorem can be used. But before using it I have to find the lengths of the diagonal. But I don't know how to. Any Hint will be helpful.


We know that $\triangle ADE \sim \triangle BCE$. Therefore, we know $|BE|=45\cdot \frac{56}{60} = 42$.

We also know $\triangle ABE \sim \triangle DCE$. Therefore $|DE|=45x$, $|CE|=42x$ and $|CD|=39x$.

Now, Ptolemy's theorem gives $60 \cdot 56 + 39 \cdot 39x = (45+42x)(42+45x)$. Solving this gives us $x=\frac7{15}$, and hence $|CD|=18.2$.

As for why this doesn't match the picture: Because your picture is inaccurate. That line with length 45 is smaller than the one with length 39.

  • $\begingroup$ How do you know $\triangle ADE$ and $\triangle BCE$ are similar? $\endgroup$ – Tebbe Feb 2 '16 at 19:17
  • $\begingroup$ @Tebbe It is just a standard fact about quadrilaterals, but if you want a proof, use the theorem that says that $\angle ADB = \angle ACB$, and $\angle ADB = \angle ACB$. I can't find its English name, unfortunately. $\endgroup$ – wythagoras Feb 2 '16 at 19:22
  • $\begingroup$ Yes, of course, inscribed angles. Thank you. $\endgroup$ – Tebbe Feb 2 '16 at 19:28

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