Question

. Prove that any point on the bisector of an angle is equidistant from
the arms of the angle
Hint: Take any point P on bisector BD of angle ABC, Draw PM perpendicular on AB and
PN perpendicular BC Then prove triangle PBM congruent triangle PBN ( by ASA criterion of congruence)
At last, prove PM = PN (as corresponding parts of congruent triangles
are equal)]​

Answer 1

Answer:

Solution-

we need to show BP=CP

in right angles △ABP

sinα= BP

AP ...(i)

in right angled △ACP

sinα= CP

AP ...(ii)

from (i) & (ii)

BP = CP

Ap AP

BP=CP

Hence Proved