# Finding a diagonal of a trapezoid that touches 3 points on a circle

In the image below: - AB and AD are tangent to the circle - BC and AD are parallel

What is the length of AC?

Thank you very much in advance!

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What have you tried? Your earlier question about a trapezoid was readily answered. It was suggested you start with cataloging observations which might be useful or give some insight. Please, in the future, you can show a little effort? – amWhy Nov 9 '12 at 21:57
Let $AB$ and $CD$ intersect at $E.$ Let $x = |EB|.$ Then, since $BC$ and $AD$ are parallel, $EBC$ and $EAD$ are similar triangles, so $\frac{EB}{EC}$ = $\frac{EA}{ED}$ = $\frac{AB}{CD} = \frac{7}{4},$ so $|EC| = \frac{4x}{7}$. By power of a point on point E, $x^2 = \frac{4x}{7}(\frac{4x}{7}+28)$, or $\frac{7x}{4} = \frac{4x}{7}+28.$ Solving, $x = \frac{784}{33}$ and $|EC| = \frac{448}{33}$. Because $AB$ and $AD$ are tangent lines, they have equal lengths, so $|AD|$ = 49. We can now apply Stewart's theorem to find the answer. This calculation is rather messy though... – only Nov 9 '12 at 21:57

Note that $AD=49$ and $BD=28$ by symmetries. Thus $ABD$ is completely determined and from there also $BC$, i.e. the complete figure.
The area of $ABD$ can be found by Heron's formula. That allows you to find the height $h$ of $B$ above $AD$, hence $\frac12 BC$ by Pythagoras from $h$ and $CD$. Finally $AC$ with Pythagoras from $AD+\frac12 BC$ and $h$.