In a convex quadrilateral
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
, the diagonal
![BD](/media/m/1/1/f/11f65a804e5c922ee28a53b1df04d138.png)
bisects neither the angle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
nor the angle
![CDA](/media/m/d/1/9/d19754e826d51d8da2971e0b887d590c.png)
. The point
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
lies inside
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
and satisfies
![\angle PBC=\angle DBA \quad \text{and} \quad \angle PDC = \angle BDA.](/media/m/5/7/a/57ab1749c546e7cfe65162bb54b5c0a4.png)
Prove that
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
is a cyclic quadrilateral if and only if
![AP=CP](/media/m/c/9/b/c9bcdd5c425926dc9f03571a1b755233.png)
.
%V0
In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies $$\angle PBC=\angle DBA \quad \text{and} \quad \angle PDC = \angle BDA.$$
Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.