Q7 of 24 Page 256

In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(∆AOD) = ar(∆BOC).

Drop perpendiculars from points D and C on segment AB as shown



Given CD || AB


Therefore the perpendicular distance between the parallel lines I equal


⇒ DG = CH = h


Consider ΔABD


Base = AB


Height = GD = h


Area(ΔABD) = × AB × h …(i)


Consider ΔABC


Base = AB


Height = CH = h


Area(ΔABC) = × AB × h …(ii)


From (i) and (ii)


Area(ΔABD) = Area(ΔABC) …(*)


Consider ΔAOD


Area(ΔAOD) = area(ΔABD) - area(ΔABO) …(iii)


Consider ΔBOC


Area(ΔBOC) = area(ΔABC) - area(ΔABO)


But Area(ΔABD) = Area(ΔABC) from (*)


⇒ Area(ΔBOC) = area(ΔABD) - area(ΔABO) …(iv)


Using (iii) and (iv)


Area(ΔAOD) = Area(ΔBOC)


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 6

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 8

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(i) ar (∆ACB) = ar (∆ACF)


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