# Application of Congruency of Triangles

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.