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