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:**

1. In ∆OPM and ∆OPN, (i) ∠MOP = ∠NOP. (ii) ∠OMP = ∠ONP = 90° (iii) OP = OP 2. ∆OPM ≅ ∆OPN. 3. PM = PM. (Proved) |
1. (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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