Q. 104.6( 9 Votes )
How will you find the point in side BC of ΔABC which is equidistant from sides AB and AC.
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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