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