# Circle Geometry-How to prove $AC \perp BD$

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 $AB/2$. Diagram

My work so far

My annotated diagram: (http)://i.stack.imgur.com/lwJcQ.png

(Sorry I cannot post more than 2 links)

So I just need to prove that $AC\perp BD$ and I can solve the problem from there. How do I do this?

• Is it? Hmm,.... – Stacker Sep 5 '16 at 19:59
• @MichaelBiro imgur.com/a/oKLYM – Jam Sep 5 '16 at 20:28

Consider the common tangent line of the two circles at the point $P$, and suppose that it insects AB at the point E. It is easy to see that the angle $\angle ACP=\angle APC=\angle EPA$ and $\angle PDB=\angle BPD=\angle EPB$. From these relations it is easy to see that $\angle ACP+\angle BDP=90$ and thus $AC\perp BD$.