In figure X and Y are the mid-points of AC and AB respectively, `QP || BC` and CYQ and BXP are straight lines. Prove that `ar (DeltaABP) = ar (DeltaACQ)`.

Explanation

Given X and Y are the mid-points of AC and AB respectively. Also, `QP || BC` and CYQ, BXP are straight lines.
To prove `ar (DeltaABP) = ar (DeltaACQ)`
Proof Since, X and Y are the mid-points of AC and AB respectively.
So, `" " XY || BC`
We know that, triangles on the same base and between the same parallels are equal in area.
Here, `DeltaBYC` and `DeltaBXC` lie on same base BC and between the same parallels BC and XY.
So, `" " ar (DeltaBYC) = ar (DeltaBXC)`
On substracting `ar (DeltaBOC)` from both sides, we get
`ar (DeltaBYC) - ar (DeltaBOC) = ar (DeltaBXC) - ar (DeltaBOC)`
`rArr" "` `ar (DeltaBOY) = ar (DeltaCOX)`
On adding `ar (DeltaXOY)` both sides, we get
`ar (DeltaBOY) + ar (DeltaXOY) = ar (DeltaCOX) + ar (DeltaXOY)`
`rArr" "` `ar (Delta BYX) = ar (DeltaCXY)" "` ...(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
`therefore" "` `ar (XYAP) = ar (YXAQ)" "` ...(ii)
On adding Eqs. (i) and (ii), we get
`ar (DeltaBYX) + ar (XYAP) = ar (DeltaCXY) + ar (YXAQ)`
`rArr" "` `ar (DeltaABP) = ar (DeltaACQ)" "` Hence proved.