ABCD is a quadrilateral. A line through D, parallel to AC meets BC produced in P. Prove that ar(triangle APD)= ar(quad. ABCD)
Asked by Sthitaprajna Mishra | 4th Feb, 2014, 08:13: PM
Consider the above figure.
Two triangles with same base and between the same parallels are of equal areas.
Now subtracting the area of triangle AOC from both the sides of the equation, we have
Please check your question.
Answered by | 6th Feb, 2014, 11:25: AM
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