In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$ .

Question

In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$.

What I Tried: Here is a picture :-

It is not hard to realise that angle-chasing is useful here, and that's exactly the same thing I did. I got the required angles in the picture as shown, but I can't seem to understand how to get $\angle BPE$ once I join the line $PE$ , if there is no suitable idea other than angle-chasing which I can use here, what should I do?

Can anyone help me? Thank You.

Edit: I become a bit stupid sometimes, hence I didn't realise $E$ is the incenter which makes $PE$ the angle bisector, implying $\angle BPE = 50^\circ$.

Answer 1

HINT: Note that $BE$ and $CE$ are two of the angle bisectors of $\Delta BPC$, so where is the third one?

Answer 2

as :E is the incenter of $\Delta PEC$ therefore $PE$ bisects the angle