Solution:-

From the given figure,

AB = AC. If PM ⊥ AB and PN ⊥ AC

We have to show that, PM x PC = PN x PB

Consider the ∆ABC,

AB = AC … [given]

∠B = ∠C

Then, consider ∆CPN and ∆BPM

∠N = ∠M … [both angles are equal to 90^o]

∠C = ∠B … [from above]

Therefore, ∆CPN ~ ∆BPM … [from AA axiom]

So, PC/PB = PN/PM

By cross multiplication we get,

PC x PM = PN x PB

Therefore, it is proved that, PM x PC = PN x PB