In parallelogram ABCD, E and F are any two points on the side AB and BC respectively. Show that
ar( △ADF)=ar( △DCE)
(hint- draw EG| |AD, FH| |AB)
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Triangle DCE and parallelogram ABCD are on the same base and between the same parallels
Therefore ar(DCE)=1/2 ar(ABCD)...(1)
Similarly
ar(ADF)=1/2 ar(ABCD)...(2)
From 1 and 2 we get
ar(ADF)=ar(DCE)
Therefore ar(DCE)=1/2 ar(ABCD)...(1)
Similarly
ar(ADF)=1/2 ar(ABCD)...(2)
From 1 and 2 we get
ar(ADF)=ar(DCE)
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