GMAT exam question based on triangles and circles

Question

Two circles touch each other externally at $P$. $APC$ and $BPD$ are drawn through $P$ to meet the two circles in $A$, $B$ and $C$, $D$ respectively. Prove that triangle $PAN$ and triangle $PCD$ are similar. Also, $AB||CD$.

What I did:

I joined $B$ and $C$.

Angle $APB$ and $DPC$ are equal. (Vertically opposite angles)

Angle $BAP$ and $CDP$ are equal. Angle subtended by the same arc are equal.

Am I right?

And how to prove if $AB||CD$?

Answer 1

Here is a helpful diagram of the situation that you are describing:

It seems to me that your proof of the similarity of the triangles is correct. Here's how you would prove that $AB || CD$:

Recall the theorem that states that alternate interior angles formed by two parallel lines and a transversal are congruent, and the converse of that theorem. Since the triangles are similar, we have $$\angle ABP=\angle CDP$$ and so if we let $\overline{BD}$ be the transversal, by the converse of the alternate interior angle theorem, the lines $\overline{AB}$ and $\overline{CD}$ are parallel.

Does that help?

Answer 2

Let $a$ be a tangent to our circles in the point $P$. It should help.