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Question
In the following 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 (ABP) = ar (ACQ).
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Solution
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 (ΔABP) = ar (ΔACQ)
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, ΔBYC and ΔBXC lie on same base BC and between the same parallels BC and XY.
So, ar (ΔBYC) = ar (ΔBXC)
On subtracting ar (ΔBOC) from both sides, we get
ar (ΔBYC) – ar (ΔBOC) = ar (ΔBXC) – ar (ΔBOC)
=» ar (ΔBOY) = ar (ΔCOX)
On adding ar (ΔXOY) both sides, we get
ar (ΔSOY) + ar (ΔXOY) = ar (ΔCOX) + ar (ΔXOY)
⇒ ar (ΔBYX) = ar (ΔCXY) ...(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
ar (XYAP) = ar (YXAQ) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ)
⇒ ar (ΔABP) = ar (ΔACQ)
Hence proved.
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