Question

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.

Answer 1

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 + <b>triangle LCD</b> + triangle ABD + <b>triangle DEK</b> + triangle ADK
= triangle BLD + <b>triangle FLB</b> + triangle ABD + <b>triangle GAK</b> + triangle ADk
= triangle FDG

Answer 2

tri FDG

Answer 3

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

Answer 4

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)

Answer 5

Jabbba jabba kappale mayathu

Answer 6

This question is wrong