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

What is the length of AC?

enter image description here

Thank you very much in advance!

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    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...2012-11-09

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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 $h$.