In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AP, show that PM × PC = PN × PB.

Considering △PNC and △PMB,

∠PNC = ∠PMB (Both are equal to 90°)

∠NCP = ∠PBM (As AB = AC)

Hence by AA axiom △PNC ~ △PMB.

Since, triangles are similar so ratio of their corresponding sides will be equal.

$\Rightarrow \dfrac{PC}{PB} = \dfrac{PN}{PM} \\[1em] \Rightarrow PC \times PM = PN \times PB.$

Hence, proved that PC × PM = PN × PB.