In the below fig. X and Y are the mid-points of AC and AB respectively, QP || BC and

CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (ΔACQ).

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#### Solution

Since x and y are the midpoint AC and AB respectively

∴ XY ll BC

Clearly, triangles BYC and BXC are on the same base BC and between the same parallels

XY and BC

∴ area (ΔBYC) = area (BXC)

⇒ area (ΔBYC) = ar (ΔBOC) = ar (ΔBXC) - ar (BOC)

⇒ ar (ΔBOY) = ar (ΔCOX)

⇒ ar ( BOY) + ar (XOY) = ar (ΔCOX) + ar (ΔXOY)

⇒ ar (ΔBXY = ar (ΔCXY)

We observe that the quadrilateral XYAP and XYAQ are on the same base XY and between

the same parallel XY and PQ.

∴ area (quad XYAP ) ar (quad XYPA) ....(2)

Adding (1) and (2), we get

ar (ΔBXY) + ar (quad XYAP) = ar (CXY) + ar (quad XYQA)

⇒ ar (ΔABP) = ar (ΔACQ)

Concept: Concept of Area

Is there an error in this question or solution?

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