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

- geometry -
**Anonymous**, Thursday, November 17, 2016 at 8:36am
In given figure X and Y are the mid point of AC and AB respectively QP parallel BC and CYQ and BXP are straight line. Prove that ar Triangle ABP =ar triangle ACQ

- geometry -
**Anonymous**, Friday, December 16, 2016 at 11:52pm
Since Y and X are the nid points of AB and AC ; by mid point theorem XY || BC. Now consider the ∆ BCQ and ∆ CBP their areas are equal cz they are on same base BC and between the same parallels

: AR(BCQ) = AR(BCP)

Let the intersection point of QC and BP be O

: add AR(AQOP) both sides.....

Therefore AR (ABP) =AR(ACQ)

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