Q. 104.6( 9 Votes )

How will you find

Answer :

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.

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