Solution:
Given: EF//GH, EI//FG and FJ//GI
To prove:
- ΔEFJ ≅ ΔHGI
- Area of parallelogram EFGH = Area of parallelogram FGIJ
Proof:
SN | Statements | Reasons |
1. |
In ΔEFJ and ΔHGI
|
|
2. | ∴ ΔEFJ ≅ ΔHGI | By A.A.S statement. |
3. | Area of ΔEFJ = area of ΔHGI | Areas of congruent triangles, from statement 2 |
4. | Area of Trapezium EFGI - ΔHGI = Area of Trapezium EFGI - area of ΔEFJ | Subtracting equal triangles from Trapezium EFGI. |
5. | ∴ Area of parallelogram EFGH = Area of parallelogram FGIJ | From statement (4), by whole part axiom. |
Hence proved.