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 ∆ABC. Draw PM | on AB and
PN | BC. Then prove ∆PBM~∆PBN (by ASA criterion of congruence).
At last, prove PM = PN (as corresponding parts of congruent triangle are equal)
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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.

Answer 2

$\large\bf{\underline\blue{Good \: Evening↝}}$

EF is the bisector of ∠BOD and ∠COA. R is a point on EF, RP⊥AB and RQ⊥CD.

In ∆ORP and ∆ORQ,

OR = OR (Common)

∠RPO = ∠RQO (90°)

∠POR = ∠QOR (OR is the bisector of ∠POQ)

∴ ∆ORP ≅ ∆ORQ (AAS congruence criterion)

⇒ PR = RQ (C.P.C.T)

Thus, the perpendiculars drawn from any point on the angle bisector of an angle, to the arms of the angle, are equal.