Q. 134.5( 4 Votes )

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Answer :

Given : ABCD is a quadrilateral in which a line through D drawn parallel to AC which meets BC produced in P.


To prove: area of (∆ABP) = area of (quad ABCD)


Proof:


Here, in the given figure


∆ACD and ∆ACP have same base and lie between same parallel line AC and DP.


According to the theorem : triangles on the same base and between same parallel lines have equal


areas.


area of (∆ACD) = area of (∆ACP) -------------1


Now, add area of (∆ABC) on both side of (1)


area of (∆ACD) + (∆ABC) = area of (∆ACP) + (∆ABC)


Area of (quad ABCD) = area of (∆ABP)


area of (∆ABP) = Area of (quad ABCD)


Hence proved


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