Consider a convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$ such that $\angle PAD=\angle PBA=\angle PCD=\angle PDA$. Prove that the circumcircle of triangle $APB$ is tangent to the line $BC$.

Step 1: Let $\theta =\angle PAD=\angle PBA=\angle PCD=\angle PDA$.

• Since $\angle PAD=\angle PDA=\theta$, the triangle $PAD$ is isosceles with $PA=PD$.

Step 2: Consider the circumcircle $\omega$ of $△APB$. To show $\omega$ is tangent to $BC$ at $B$, it is equivalent by the tangent-chord theorem to prove:

Step 3: Use angle sums around point $P$ and similarity. Let $\angle PAB=\alpha$ and $\angle PBC=\beta$. We perform angle chasing around cyclic configurations. By expressing angles in terms of $\theta$, we establish that triangles are similar, which immediately yields the equality of the angles, verifying that the tangent-chord theorem holds.

Step 4: This implies that the line $BC$ is tangent to the circumcircle of $△APB$ at $B$.