Degenerate (Tangent) Converse of Power of a Point Theorem

Assume points $P, A, B, C$ are given as shown in the diagram.

Recall that the Power of a Point Theorem tells us that $(PA)^2=(PB)(PC)$ in this configuration.

Now let's prove the converse of this statement. Let $A, B, C$ be points on a given circle $(O)$, and let $P$ be on line $BC$ such that $(PA)^2=(PB)(PC)$. Let the line $PA$ intersect $(O)$ a second time at $A_1$.

According to the Power of a Point Theorem, $(PA)(PA_1)=(PB)(PC)$. But $(PA)^2=(PB)(PC)$, so $(PA)(PA_1)=(PA)^2$, and it follows that $PA=PA_1$. Since $P$ is outside of the circle, $A$ and $A_1$ are on the same side of $P$. This means that $A=A_1$, and so $PA$ is tangent to the circle.

In summary, the converse of the Power of a Point Theorem in the degenerate case tells us the following:

If $A, B, C$ are points on a given circle $(O)$, and $P$ is on line $BC$ such that $(PA)^2=(PB)(PC)$, then $PA$ is tangent to $(O)$.