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Find $MP$ 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:
![image](https://i.imgur.com/GvR89Dj_d.webp?maxwidth=760&fidelity=grand)

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<summary>My attempts:</summary>
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](https://en.wikipedia.org/wiki/Intersecting_secants_theorem) with the legendary [Pythagorean theorem](https://en.wikipedia.org/wiki/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 collinear–which I haven't yet figured out how to prove.
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Is there a way to solve this problem that doesn't involve _headache-inducing algebra crunching_?