More problems

Problem 1. The tangents at two points B and C on a circle meet at A. Let A₁B₁C₁ be the pedal triangle of the (isosceles) triangle ABC for an arbitrary point P on the circle, as in Figure 5. Proof that

PA²₁=PB₁ PC₁.

  

Figure 5: Problem 1

Hint. After dividing the both sides of the above identity by PB.PC we can see that it is sufficient to prove that $\sin \angle PBC \sin \angle PCB=\sin \angle PCB_1 \sin \angle PBC_1$, which follows by the fact that $\angle PCB=\angle PBC_1$ and $\angle PBC=\angle PCB_1$.

Problem 2. If a point P lies on the arc CD of the circumcircle of a square ABCD, proof that

PA(PA+PC)=PB(PB+PD).

Hint. Use the Ptolemy's theorem for the quadrilaterals ABCP and ACPD.