Let $ABC$ be a triangle. Let $P$ be a point on the segment $BC$. Prove that the circumcircle of $△ABP$ is tangent to $AC$ if and only if $A{P}^2=BP\cdot CP$.

Step 1: By the tangent-chord theorem, the circumcircle of $△ABP$ is tangent to $AC$ at $A$ if and only if:

Step 2: Consider the two triangles $△PAC$ and $△PBA$. Since they share the angle at $C$ (or by analyzing angle relations), if $\angle PAC=\angle PBA$, then the triangles $△PAC$ and $△PBA$ are similar. This similarity leads directly to the ratio relation of the sides:

Step 3: This shows both directions of the equivalence, completing the proof.