Question

prove that any point on the bisector of an angle is equidistant from the arms of the angles
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Answer 1

Answer:

Let ABC be an angle and let BD be its bisector. Take a point P on BD. Draw PQ and PR perpendicular to the arms AB and BC of the angle.

In ∆s BPQ and BPR

Angle PBQ = Angle PBR (since BP bisects angle B)

BP = BP (common side)

Angle BQP = Angle BRP (each = 90°)

Hence ∆BPQ is congruent to ∆BPR.

Therefore, PQ = PR (c.p.c.t.)

Hence the bisector of an angle is equidistant from the arms of the angle.

Hope this help you ;

Bhavy Yadav