$\angle BAC=25$, $\angle BCA=20$, $\angle BDC = 50$, $\angle BDA = 40$, $AC$ and $BD$ intersect at $P$, what is $\angle CPD$?

Question

In convex quadrilateral $ABCD$, $\angle BAC=25$, $\angle BCA=20$, $\angle BDC = 50$, $\angle BDA = 40$, $AC$ and $BD$ intersect at $P$, what is $\angle CPD$?

I don't really know how to start on this question. I tried letting $\angle CPD=x$ and angle chasing, but it didn't get me anywhere. I also tried extending line $CB$ to form a right triangle, but it also didn't help me. Could someone give me a hint to get me started?

Answer 1

Let $E\in BD$ such that $B$ is placed between $E$ and $P$ and $AD$ is a bisector of $\Delta EAP$.

Thus, since $\measuredangle EAC=\measuredangle EDC$, we see that $AECD$ is cyclic, which says: $$\measuredangle ECA=\measuredangle EDA=40^{\circ},$$ which gives $CB$ is a bisector of $\Delta AEC$ and from here $EP$ is a bisector of $\Delta AEC.$

Now, let $\measuredangle CPD=x$.

Thus, $$\measuredangle AEP=\measuredangle CEP=x-40^{\circ},$$ which gives: $$\measuredangle APD=x-40^{\circ}+50^{\circ}=x+10^{\circ}.$$ Id est, $$x+x+10^{\circ}=180^{\circ}$$ and $$x=85^{\circ}.$$

Answer 2

Sketch explaining solution below