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

A pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

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

- geometry -
**ar(pentagon ABCDE)=ar(triDGF)**, Saturday, October 22, 2016 at 1:36am
tri FDG

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