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September 16, 2014

September 16, 2014

Posted by **Anonymous** on Saturday, January 15, 2011 at 2:46am.

- geometry -
**Reiny**, Saturday, January 15, 2011 at 9:11amLets 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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