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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