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I'm working through some exercises in preparation for midterms and I'm stuck on the following exercise.

For ABCD a convex quadrilateral with $AB = 13 = BD$, $BC = CD$, $AD = 10$ $\measuredangle BCD =90$ Find the length of CM where M is the midpoint of AD.

Here is my attempt at the solution:

The quadrilateral ABCD

Since $\Delta BCD$ is a right triangle we may apply Pythagoras:

Let $a = BC$ , since by assumption $BC = CD$ we have $a=CD$. Then $a^{2}+a^{2} = 13^{2}$ which implies that $a =\frac{13}{\sqrt{2}}$

I want to use Apollonius' Theorem $AC^{2} + DC^{2} = 2AM^{2} + 2CM^{2}$

But I'm having trouble coming up with a strategy to finding the length of AC. Since once AC is found I may use Apollonius' Theorem to find the length AM.

Any hints on how to proceed would be greatly appreciated. Thanks.

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    Is your main goal finding $AC$.? – 2017-02-19
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    Yeah, since once AC is found, Apollonius theorem will provide the desired length CM. – 2017-02-19
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    Is cosine relation allowed.? – 2017-02-19
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    **Perhapse helps:** $BM\perp AD$ then $BM=\sqrt{169-25}=12$. $\angle B_{down}=\angle D_{right}=45^\circ$, from cosine relation $$BM^2=MD^2+BD^2-2DM.BD\cos\angle D_{up}$$ you will find $\cos\angle D_{up}=\dfrac{5}{13}$. We have $CM^2=DM^2+DC^2-2DM.DC\cos(45^\circ+D_{up})$. – 2017-02-19
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    Unfortunately not. Only basic congruency theorems, Thales theorem, and similar triangles. – 2017-02-19

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Since $\measuredangle BMD = \measuredangle BCD = 90^{\circ}$, $BCDM$ is cyclic. Thus we can use Ptolemy's theorem: $$BM\cdot CD + MD\cdot BC = MC\cdot BD,$$ $$12\cdot\frac{13}{\sqrt{2}}+5\cdot\frac{13}{\sqrt{2}}=MC\cdot 13,$$ $$MC=\frac{17}{\sqrt{2}}.$$