# In Fig. 15.87, X and Y are the mid-point of AC and AB respectively, QP||BC and CYQ and BXP are straight lines. Prove that:ar(Δ ABP) = ar(Δ ACQ).

In a ΔAXP and ΔCXB,

PAX = XCB (Alternative angles AP || BC)

AX = CX (Given)

AXP = CXB (Vertically opposite angles)

ΔAXP ΔCXB (By ASA rule)

AP = BC (By c.p.c.t) (i)

Similarly,

QA = BC (ii)

From (i) and (ii), we get

AP = QA

Now,

AP || BC

And,

AP = QA

Area (APB) = Area (ACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)

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