Here we will prove that the bisectors of the angles of a triangle meet at a point.
Solution:
Given In ∆XYZ, XO and YO bisect ∠YXZ and ∠XYZ respectively.
To prove: OZ bisects ∠XZY.
Construction: Draw OA ⊥ YZ, OB ⊥ XZ and OC ⊥ XY.
Proof:
Statement 1. In ∆XOC and ∆XOB, (i) ∠CXO = ∠BXO (ii) ∠XCO = XBO = 90° (iii) XO = XO.
2. ∆XOC ≅ ∆XOB 3. OC = OB 4. Similarly, ∆YOC ≅ ∆YOA 5. OC = OA 6. OB = OA. 7. In ∆ZOA and ∆ZOB, (i) OA = OB (ii) OZ = OZ (iii) ∠ZAO = ∠ZBO = 90 8. ∆ZOA ≅ ∆ZOB. 9. ∠ZOA = ∠ZOB. 10. NO bisects ∠XZY. (Proved) 
Reason 1. (i) XO bisects ∠YXZ (ii) Construction. (iii) Common Side.
2. By AAS criterion of congruency. 3. CPCTC. 4. Proceeding as above. 5. CPCTC. 6. Using statement 3 and 5. 7. (i) From Statement 6. (ii) Common Side. (iii) Construction.
8. By RHS criterion of congruency. 9. CPCTC. 10. From statement 9. 
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