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. |
From Bisectors of the Angles of a Triangle Meet at a Point to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Apr 23, 24 04:50 PM
Apr 23, 24 03:15 PM
Apr 22, 24 01:35 PM
Apr 21, 24 10:57 AM
Apr 20, 24 05:39 PM