Mathematics
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)
Theorems on Area
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Answer
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).
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