Given ABCD is a parallelogram and E is a point on BC. AE and DC are produced to meet at F.
`therefore" "` `AB || CD "and" BC || AD" "` …(i)
To prove `ar(DeltaADF) = ar(ABFC)`
Construction Join AC and DE.
Proof Since, AC is a diagonal of parallelogram ABCD.
So, `" " ar (DeltaABC) = ar (DeltaACD)" "` ...(ii)
Since, `DeltaABF` and `DeltaABC` are on the same base AB and between the same parallels AB and DF. `" "` [since, `AB || DC` and DC produced to F ]
`therefore" "` `ar(DeltaABF) = ar (DeltaABC)" "` ...(iii)
From Eqs. (ii) and (iii).
`ar (DeltaABC) = ar (DeltaACD) = ar (DeltaABF)`
On subtracting `ar (DeltaABE)` from both sides of Eq. (iii), we get
`ar (DeltaABF) - ar (DeltaABE) = ar (DeltaABC)- ar (DeltaABE)`
`rArr" "` `ar (DeltaBEF) = ar (DeltaAEC)" "` ...(iv)
Now, `" " ar (AECD) = ar (ACD) + ar (AEC)`
`= ar (DeltaABC) + ar (DeltaBEF)" "` [from Eqs. (ii) and (iv)]
On adding `ar (DeltaCEF)` both sides, we get
`ar (AECD) + ar (DeltaCEF)`
`= ar (DeltaABC) + ar (DeltaBEF) + ar (DeltaCEF)`
`rArr" "` `ar (DeltaADF) = ar (ABFC)" "` Hence proved.
