Given a quadrangle $ABCD$ where two of the sides $AD$ and $BC$ are parallel with lengths $a$ and $b$ respectively.
Futhermore we have the midpoint $M$ on $AB$ and $K$ on $CD$. The diagonals $AC$ and $BD$ intersects the line $KM$ at points $P$ and $Q$ respectively. I am tasked with expressing the distance $PQ$ in terms of $a$ and $b$.

This is problem in Euclidean geometry I've been stuck on for quite a while. So far I've narrowed it down to two cases: one where we get a trapezoid like figure and one where we the angles at $B$ and $D$ are greater than 90 degrees. 
I did two quick figures in paint (I apologize for my artistic qualities beforehand) to illustrates this.

![](https://i.stack.imgur.com/KTwbg.png)
![](https://i.stack.imgur.com/kLjrY.png)

So far I've been able to show that the two triangles formed by the diagonals and $AD$ and $BC$ are similar but that is a far as I'm able to get.

My current idea is to try to show that $KM$ is parallel with either of the two parallel line. Then the triangle formed by the diagonals and their intersection with $KM$ will be similar to both the other two triangles I found. This should enable me to find an expression for $PQ$

However I've been unable to find a way to show this.