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.
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.