In the figure, M is the mid point of AE. Prove that the area of triangle ΔABE is equal to the area of parallelogram ABCD.
1 Answer
Solution:
Given:
M is the mid point of AE and ABCD is a parallelogram
To prove: Area of triangle ΔABE = area of parallelogram ABCD
Construction: BM is joined.
Proof:
| Statements | Reasons | |
| 1. | area of triangle ΔAMB = 1/2 area of parallelogram ABCD Or, Area of parallelogram ABCD = 2 x area of triangle ΔAMB |
Both are standing on the same base and between the same parallel lines. |
| 2. | area of triangle ΔAMB = 1/2 area of triangle ΔABE Or, area of triangle ΔABE = 2 x area of triangle ΔAMB |
Median BM bisects triangle ΔABE |
| 3. | area of triangle ΔABE = area of parallelogram ABCE | from statements 1 and 2, being double of the same area triangle ΔAMB |
| 4. | ∴ Area of triangle ΔABE = area of parallelogram ABCD | from statement 3 |
Hence proved
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