Question:
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Draw FG ⊥ AB
We have,
BE = 2 EA and DF = 2FC
⇒ AB – AE = 2 AE and DC – FC = 2 FC
⇒ AB = 3 AE and DC = 3 FC
⇒ AE = (1/3) AB and FC = (1/3)DC ⋅⋅⋅⋅ (1)
But AB = DC
Then, AE = FC [opposite sides of ∥gm]
Thus, AE = FC and AE ∥ FC
Then, AECF is a parallelogram
Now, area of parallelogram AECF = AE × FG
⇒ ar(∥gm AECF) = 1/3 AB × FG from(1)
⇒ 3ar (∥gm AECF) = AB × FG ⋅⋅⋅ (2)
And ar(∥gm ABCD) = AB × FG ⋅⋅⋅ (3)
Compare equation 2 and 3
⇒ 3ar(∥gm AECF) = ar(∥gm ABCD)
⇒ ar(∥gm AECF) = 1/3 ar(∥gm ABCD)
