# <span lang="EN-US

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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