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.
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.
(iii) Common side.
2. By AAS criterion.
Note: The abbreviation CPCTC is generally used for ‘Corresponding parts of Congruent Triangles are Congruent’.
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