In the given figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that: ar (△ABP) = ar (quad.ABCD)

Triangles △ ACD and △ ACP on the same base AC and between same parallels AC and DP.

∴ ar (△ACD) = ar (△ACP)

Add the common ar(△ABC) to both sides

ar (△ACD) + ar(△ABC) = ar (△ACP) + ar(△ABC)

From figure,

∴ ar (quad.ABCD) = ar (△ABP)

Hence, proved that ar (△ABP) = ar (quad. ABCD).