Prove that $\angle BAC + \angle OAP = 180^\circ$

Question

Prove that if you construct two circle centered at O and P and intersecting at A with tangent lines BA and CA. Prove that $\angle BAC + \angle OAP = 180^\circ$.

I'm having trouble just starting the proof.

Answer 1

Hint: Look at the other intersection $D$ of the circles. The quadrilateral $OAPD$ is symmetric about $OP$. Now we only need to show $\angle BAC$ is one half of the sum of $\angle DOA$ and $\angle DPA$.

Answer 2

Use the fact that $OA \perp BA, PA \perp AC$. In directed angles:

$\angle BAC = \angle OAP$

So now it just matters about which orientation you will use, because if you take $\angle OAP$ as to be in the opposite direction to $\angle BAC$, then you will obtain the desired expression.