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While drawing the figure we can see that AR(DAE)=1/3AR(ABD). Now AR(ABD)=1/2AR(ABCD). By substituting this, we get AR(DAE)= 1/3×1/2AR(ABCD)= 1/6AR(ABCD)
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Join D to E,F and B.we will get three different triangles.now DAB and ABCD are on the same base and between the same parallels.so AreaDAB =1/2areaABCD. in the same way DAF would be equal to1/2 AreaDAB. so indirectly DAF would be equal to1/4 areaABCD.. then areaDAE would be equal to 1/2 area DAF. Indirectly area DAE would be equal to 1/6 area ABCD....Hence proved that areaDAE=1/6areaABCD.
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