Question

point p is the equidistant from the arms of the angle abc prove that BP bisects angle ABC​

Answer 1

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.

Step-by-step explanation:

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Answer 2

Answer:

Consider the image attached.

• Draw PQ and PR perpendicular to the arms AB and BC of the angle.

• Then, we have joined BP.

We, have to show that BP is the bisector of ∠ABC.

• According to the given problem , point P is the equidistant from the arms of the ∠ABC, So , PQ = PR

Consider ΔQBP and ΔRBP,

• PQ = PR (given , point p is the equidistant from the arms of the angle ABC)

• ∠PQB = ∠PRB (Both are 90°)

• PB is the common sides for Both the Triangle.

Hence using RHS rules of Congruency, we can say ΔQBP ≅ ΔRBP

So, ∠QBP =∠RBP [from CPCT rules]

Hence, we can say that BP is a bisector of ∠QBR as well as ∠ABC.

Here it is proved that if point P is the equidistant from the arms of the angle ABC , then BP bisects angle ∠ABC​.

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