In the given figure, AD//BE//GF and AB//DG//EF. If the areas of parallelogram ABCD and parallelogram CEFG are equal, then prove that DE//BG.
1 Answer
Given: AD//BE//GF, AB//DG//EF
and parallelogram ABCD = Parallelogram CEFG
To prove: DE//BG
Construction: BD and GE are joined.
Proof:
| Sn | Statements | Reasons |
| 1 | Triangle BCD = 1/2 parallelogram ABCD | Diagonal of a parallelogram bisects it. |
| 2 | Triangle GCE = 1/2 parallelogram CEFG | Same reason as (1). |
| 3 | Parallelogram ABCD = Parallelogram CEFG | Given |
| 4 | Triangle BCD = Triangle GCE | From statements 1, 2, and 3, both are halves of the equal parallelogram |
| 5 | Triangle BCD + Triangle BCG = Triangle GCE + Triangle BCG | Adding triangle GCE on both sides of statement 4. |
| 6 | Triangle BDG = Triangle BEG | From statement 5, by whole part axiom. |
| 7 | Therefore, DE//BG | From statements 6, being triangles on the same like segment and towards the same side of it equal in area. |
Proved
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