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 = OP 2. ∆OPM ≅ ∆OPN. 3. PM = PM. (Proved) |
Reason 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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