# 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.

To prove PM = PN.

Proof:

 Statement 1. In ∆OPM and ∆OPN,(i) ∠MOP = ∠NOP.(ii) ∠OMP = ∠ONP = 90°(iii) OP = OP2. ∆OPM ≅ ∆OPN. 3. PM = PM. (Proved) Reason1. (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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