Posted by Anonymous on Saturday, January 15, 2011 at 2:46am.
Lets concentrate on the quadrilateral ADEG.
let the intersection of AE and DB be K
GE || AD (given)
So triangles ADE and ADG have the same base AD and the same height , since parallel lines are equal distances apart.
So triangle ADE = triangle ADG
but ADK is common to both, so subtracting ...
triangle AKG = triangle DEK
similarly by letting FD and BC intersect at L we can show that triangle FLB = triangle CDL
pentagon ABCDE
= triangle BLD + triangle LCD + triangle ABD + triangle DEK + triangle ADK
= triangle BLD + triangle FLB + triangle ABD + triangle GAK + triangle ADk
= triangle FDG
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