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ABCD is a parallelogram. E is a point on BA such that BE=2FA and F is a point on DC such that DF=2FC. prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

Asked by manavmansvi 7th March 2015, 11:38 AM
Answered by Expert
Answer:
L e t space t h e space h e i g h t space o f space t h e space p a r a l l e log r a m space b e space apostrophe h apostrophe I n space p a r a l l e log r a m space A B C D comma D C equals D F plus F C equals 2 F C plus F C equals 3 F C A r e a space o f space p a r a l l e log r a m space A B C D equals h cross times C D equals h cross times 3 F C A r e a space o f space p a r a l l e log r a m space A E C F space equals h space cross times F C fraction numerator A r e a space o f space p a r a l l e log r a m space A B C D over denominator A r e a space o f space p a r a l l e log r a m space A E C F end fraction equals fraction numerator h cross times 3 F C over denominator h space cross times F C end fraction fraction numerator A r e a space o f space p a r a l l e log r a m space A B C D over denominator A r e a space o f space p a r a l l e log r a m space A E C F end fraction equals 3 over 1 A r e a space o f space p a r a l l e log r a m space A E C F equals 1 third A r e a space o f space p a r a l l e log r a m space A B C D H e n c e space P r o v e d
Answered by Expert 7th March 2015, 11:59 PM
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