D and F are point on side ACof triangle ABE through point D a line DC is drawn which is parallel to AB and intersect BE at C prove that area of triangle ACF equal to area of BCFD
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We know that triangle having common base
and between same parallels are equal in area.Since ∆ABC and ∆ABD have common base AB and between same parallel lines AB and DC, then
ar(∆ABC) = ar(∆ABD) ......(1)Subtracting ar(∆AOB) from both sides, we get ar(∆ABC) − ar(∆AOB) = ar(∆ABD) − ar(∆AOB) ⇒ar(∆OBC) = ar(∆AOD)Adding ar(□DOCF) both sides of the above equation, we get ar(∆OBC) + ar(□DOCF) = ar(∆AOD) + ar(□DOCF)⇒ar(□BCFD) = ar(∆ACF)
and between same parallels are equal in area.Since ∆ABC and ∆ABD have common base AB and between same parallel lines AB and DC, then
ar(∆ABC) = ar(∆ABD) ......(1)Subtracting ar(∆AOB) from both sides, we get ar(∆ABC) − ar(∆AOB) = ar(∆ABD) − ar(∆AOB) ⇒ar(∆OBC) = ar(∆AOD)Adding ar(□DOCF) both sides of the above equation, we get ar(∆OBC) + ar(□DOCF) = ar(∆AOD) + ar(□DOCF)⇒ar(□BCFD) = ar(∆ACF)
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