Prove for Pedal & Isosceles triangle.

Question

The tangents at two points $B$ and $C$ on a circle meet at $A$. Let $A_1B_1C_1$ be the pedal triangle of the isosceles triangle $ABC$ for an arbitrary point $P$ on the circle, as shown below. Then prove that $${PA_1}^2 = PB_1 \times PC_1$$

Answer 1

Hint: The triangles $PC_1A_1$ and $PA_1B_1$ are similar (both are similar to the triangle $PBC$).

Answer 2

Another hint: Note that $PA_1BC_1$ has 2 opposite $90^\circ$ angles. Hence it is cyclic. This can help get some angles to prove that $PC_1A_1$ is similar to $PBC$.

Answer 3

$P C_1 B A_1$ is a cyclic quadrilateral since the angles at $A_1$ and $C_1$ are right angles, so $\widehat{PC_1 A_1}=\widehat{P B_1 A_1}$. Moreover, $$ \widehat{C_1 P A_1} = \pi-\widehat{C_1 B A_1} = \widehat{CBA} = \widehat{BPC} $$ so $PC_1A_1$ and $PBC$ are similar triangles. In the same way, $PA_1 B_1$ and $PBC$ are similar triangles.

You may also apply a circular inversion centered at $P$ and see what happens.