# Finding side lengths of a trapezoid given the distance between its diagonal intersection and midpoint of a diagonal

## 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?

• Why not tell your "friend" you are unable to prove it, and suggest your "friend" create an account and post his/her question here? Feb 4 at 22:17
• Well, I'm personally interested in finding out more about the problem, so I decided to post it here. I've already spent a bit of time and want to see a solution myself. Feb 4 at 22:17
• So then it's your own question. I see not need to cite a "friend", because you it has become your question, right? Feb 4 at 22:19
• So what's the given length of $MI$? Without it, there is insufficient information. Feb 4 at 22:20
• Oh ok, I thought I saw something about putting the origin of the question. Sorry about that, I'm new here as well so not 100% sure what's accepted and what's not. Feb 4 at 22:20

$$CM // AD$$. 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 $$AD=2CM$$.
• So just checking, would $4\sqrt{190}$ would be the correct answer? Feb 4 at 22:57