Point on the Bisector of an Angle

Here we will prove that any point on the bisector of an angle is equidistant from the arms of that angle.

Solution:

Given OZ bisects ∠XOY and PM ⊥ XO and PN ⊥ OY.

$Point on the Bisector of an Angle$

To prove PM = PN.

Proof:

Statement

1. In ∆OPM and ∆OPN,

(i) ∠MOP = ∠NOP.

(ii) ∠OMP = ∠ONP = 90°

(iii) OP = OP

2. ∆OPM ≅ ∆OPN.

3. PM = PM. (Proved)

Reason

(i) OZ bisects ∠XOY.

(ii) Given

(iii) Common side.

2. By AAS criterion.

3. CPCTC.

Note: The abbreviation CPCTC is generally used for ‘Corresponding parts of Congruent Triangles are Congruent’.

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