How will you find

In Δ ABC, draw bisector of BC which cuts it at D.

Take any point P on AD. Draw PN⊥ AB and PM ⊥ AC.

In Δ APN and Δ APM

∠ PNA = ∠ PMA = 90° ( by construction)

∠PAN = ∠PAM ( AD is bisector of ∠A)

AP = AP )( common)

Δ APN ≅ Δ APM (By AAS rule)

⇒ PN = PM ( By CPCT)

So, P is equidistant from AB and AC.

∴ any point on AD will be equidistant from AB and AC.