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.


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

Point on the Bisector of an Angle

To prove PM = PN.



1. In ∆OPM and ∆OPN,

(i) ∠MOP = ∠NOP.

(ii) ∠OMP = ∠ONP = 90°

(iii) OP = OP

2. ∆OPM ≅ ∆OPN.

3. PM = PM. (Proved)



(i) OZ bisects ∠XOY.

(ii) Given

(iii) Common side.

2. By AAS criterion.


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

9th Grade Math

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