# Find length $MP$ in trapezoid

Trapezoid $ABCD$ is given such that $AB\parallel CD$ and $AD=BD$. Let $M$ be the midpoint of $AB$ and $P$ be the intersection of diagonal $AC$ with the circumcircle of $\triangle BCD$. Given that $BC=27$, $CD=25$, and $AP=10$, find $MP$.

Here's some visualization that I made in GeoGebra:

## My attempts:

The answer key states the answer is $\frac{27}5$, which I found to be true using GeoGebra.To actually solve it, I later named the trapezoid's height as $h$, extended $AB$ to intersect the circle again at $Q$, named a certain length $d$, used the intersecting secants theorem with the legendary Pythagorean theorem that we all know, went through some *heachache-inducing algebra crunching*, and eventually got it with the help of a calculator.

But, I thought, there must be a way to solve this that isn't so egregiously unelegant! Right?

So I named some angles. $\alpha=\angle BAC=\angle ACD=\angle PBD$, $\beta=\angle ABP=\angle CAD$, $\gamma=\angle ACB=\angle PDB$, $\delta=\angle BDC=\angle BPC$, $\epsilon=\angle CBD=\angle CPD$, $\zeta=\angle MDP$.

But I couldn't find much similarity that would be of use. Only that $\triangle APB\sim\triangle BDP\sim\triangle ACD$, which leads to $\delta=\alpha+\beta$ and $AB=\frac25AC$.

Even more mysteriously, with GeoGebra I was able to find that $\angle MPB=\epsilon$! How is that, and would that be meaningful information? I even tried extending $BC$ to the point $F$ such that $\triangle ABC\sim\triangle CFP$ (i.e. $AFBP$ a cyclic quadrilateral), which would require $F,M,P$ to be collinearâ€“which I haven't yet figured out how to prove.

Is there a way to solve this problem that doesn't involve *headache-inducing algebra crunching*?

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