Q. 255.0( 3 Votes )

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).


Answer :

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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