Q10 of 28 Page 112

How will you find the point in side BC of ΔABC which is equidistant from sides AB and AC.



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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