In figure ABCD is a quadrilateral and BEparallelAC and also BE meets DC produced at E show that area of ADE is equal to the area of the quadirilateral ABCD
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∆BAC and ∆EAC lie on the same base AC and are between the same parallel lines AC and BE
Therefore, area(∆BAC) = area(∆EAC) [by theorem 9.2 (2 triangles in the same base and between the same parallel lines are equal in area) ]
So, area(∆BAC) + area(∆ADC) = area(∆EAC) + area(∆ADC) [Adding same areas on both sides]
area(ABCD) = area(∆ADE)
Therefore, area(∆BAC) = area(∆EAC) [by theorem 9.2 (2 triangles in the same base and between the same parallel lines are equal in area) ]
So, area(∆BAC) + area(∆ADC) = area(∆EAC) + area(∆ADC) [Adding same areas on both sides]
area(ABCD) = area(∆ADE)
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