Two circles are externally tangent at point $P$, as shown. Segment $\overline{CPD}$ is parallel to common external tangent $\overline{AB}$. Prove that the distance between the midpoints of $\overline{AB}$ and $\overline{CD}$ is $AB/2$.
I first started with some construction: We start by drawing a common tangent to the two circles passing through $P$ and intersecting $AB$ at $M$. Then we extend $AC$ and $BD$ so that they intersect at $Q$.Next, we extend $QM$ to meet $CD$ at N.
I divided the proof into five steps: 1)$M$ is midpoint of $AB$. 2)$M$ is the circumcenter of $\triangle APB$. 3)$A$ and $B$ are the midpoints of $\overline{CQ}$ and $\overline{DQ}$, respectively. 4) $N$ is the midpoint of $\overline{CD}$. 5)$MN=MP$.
I could prove (1) through (4) and I am not able to prove (5). Can anyone help me out with this?