geometryquadrilateral
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.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. So they are necessarily equidistant from the point $O$, aren't they?
And you have two arcs: $CD=120$, and $BC=60$. So, $BD=180$.
Best Answer
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.