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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