# geometry

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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 - ,

let the intersection of AE and DB be K
So triangles ADE and ADG have the same base AD and the same height , since parallel lines are equal distances apart.
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 - ,

tri FDG

• geometry - ,

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 - ,

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
Therefore AR (ABP) =AR(ACQ)

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Jabbba jabba kappale mayathu

• geometry - ,

This question is wrong