The Question
My friend recently gave me a problem that I was interested in, but could not completely solve. I've already constructed a diagram that roughly represents the problem, and it's below.
Consider a trapezoid $ABCD$ with $AB \parallel CD$. Additionally, side $BC = CD = 43$ and $AD \perp BD$. Given that the length between the midpoint of diagonal $BD$ and intersectional of diagonals $AC$ and $BD$ is $11$, find the side length $AD$. Express your answer in simplified radical form.
My Understanding
So my first idea was to draw it out, and I've managed to get a good approximation of it.
So I gave some names to the other points, calling the midpoint of the diagonal $M$ and the intersection $I$. The first thing that I noted was that because $BC = CD$, I found that drawing it down to the midpoint of diagonal $BD$ created two right triangles. Additionally, I noted congruent angles within the triangle $CPO$ and triangle $ADO$, so I deduced they were similar. But I'm not sure how to proceed from here, as I'm not able to establish the side lengths of any more segments. Does anyone know how to proceed?
Best Answer
$CM // AD$. Extend it to meet $AB$ in $E$. Alternate interior angles $\angle MBE=\angle MDC$ which is equal to $\angle MBC$. Thus $\triangle MBC \cong \triangle MBE$. So $M$ bisects $CE$.
$AECD$ is a parallelogram. So $AD=CE$. We conclude $\triangle ADI \sim \triangle CMI$ with similarity ratio $2$.
Thus $MI=11\Rightarrow ID=22 \Rightarrow MD=33$.
From Pythagoras, you can find $CM$ and $AD=2CM$.