Here we will prove some Application of congruency of triangles.
1. PQRS is a rectangle and POQ an equilateral triangle. Prove that SRO is an isosceles triangle.
Solution:
Given:
PQRS is a rectangle. POQ is an equilateral triangle to prove ∆SOR is an isosceles triangle.
Proof:
Statement 
Reason 
1. ∠SPQ = 90° 
1. Each angle of a rectangle is 90° 
2. ∠OPQ = 60° 
2. Each angle of an equilateral triangle is 60° 
3. ∠SPO = ∠SPQ  ∠OPQ = 90°  60° = 30° 
3. Using statements 1 and 2. 
4. Similarly, ∠RQO = 30° 
4. Proceeding as above. 
5. In ∆POS and ∆QOR, (i) PO = QO (ii) PS = QR (iii) ∠SPO = ∠RQO = 30° 
5. (i) Sides of an equilateral triangle are equal. (ii) Opposite sides of a rectangle are equal. (iii) From statements 3 and 4. 
6. ∆POS ≅ ∆QOR 
6. By SAS criterion of congruency. 
7. SO = RO 
7. CPCTC. 
8. ∆SOR is an isosceles triangle. (Proved) 
8. From statement 7. 
2. In the given figure, triangle XYZ is a right angled at Y. XMNZ and YOPZ are squares. Prove that XP = YN.
Solution:
Given:
In ∆XYZ, ∠Y = 90°, XMNZ and YOPZ are squares.
To prove: XP = YN
Proof:
Statement 
Reason 
1. ∠XZN = 90° 
1. Angle of square XMNZ. 
2. ∠YZN = ∠YZX + ∠XZN = x° + 90° 
2. Using statement 1. 
3. ∠YZP = 90° 
3. Angle of square YOPZ. 
4. ∠XZP = ∠XZY + ∠YZP = x° + 90° 
4. Using statement 3. 
5. In ∆XZP and ∆YZN, (i) ∠XZP = ∠YZN (ii) ZP = YZ (iii) XZ = ZN 
5. (i) Using statements 2 and 4. (ii) Sides of square YOPZ. (iii) Sides of square XMNZ. 
6. ∆XZP ≅ ∆YZN 
6. By SAS criterion of congruency. 
7. XP = YN. (Proved) 
7. CPCTC. 
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