Question

In Fig. 9.26, 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).

Answer 1

As X and Y are midpoint of AC and AB respectively 
so XY || BC

the triangle BYC and BXC are on the same base BC and between the same parallels XY and BC

ar (BYC) = ar (BXC)

ar (BYC) - BOC = ar (BXC) - triangle BOC

ar (BOY) = ar (COX)

ar (BOY) + XOY = ar (COX) + XOY

ar (BXY) = ar (CXY)  --- (1)

ar (XYAP) = ar (XYQA) --- (2)

(1) + (2)

ar (ABP) = ar (ACQ)

Answer 2

Answer:

Step-by-step explanation:

As X and Y are midpoint of AC and AB respectively 

so XY || BC

the triangle BYC and BXC are on the same base BC and between the same parallels XY and BC

ar (BYC) = ar (BXC)

ar (BYC) - BOC = ar (BXC) - triangle BOC

ar (BOY) = ar (COX)

ar (BOY) + XOY = ar (COX) + XOY

ar (BXY) = ar (CXY)  --- (1)

ar (XYAP) = ar (XYQA) --- (2)

(1) + (2)

ar (ABP) = ar (ACQ)

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