Question

D AND F ARE POINTS ON SIDE AE OF TRIANGLE ABE. THROUGH POINT D A LINE DC IS DRAWN WHICH IS PARALLEL TO AB AND MEETS BE IN C. PROVE THAT ar(ACF)=ar(BCFD)

Answer 1

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, thenar(∆ABC) = ar(∆ABD) ......(1)Subtracting ar(∆AOB) from both sides, we getar(∆ABC) − ar(∆AOB) = ar(∆ABD) − ar(∆AOB) ⇒ar(∆OBC) = ar(∆AOD)Adding ar(□DOCF) both sides of the above equation, we getar(∆OBC) + ar(□DOCF) = ar(∆AOD) + ar(□DOCF)⇒ar(□BCFD) = ar(∆ACF)