Question -
Answer -
Given:
ABCD is a parallelogram
G is a point such that AG = 2GB
E is a point such that CE = 2DE
F is a point such that BF = 2FC
To prove:
(i)
(ii)
(iii)
(iv)
What portion of the area of parallelogram ABCD is the area of ΔEFG
Construction: draw a parallel line to AB through point F and a perpendicular line to AB through
PROOF:
(i) Since ABCD is a parallelogram,
So AB = CD and AD = BC
Consider the two trapeziums ADEG and GBCE:
Since AB = DC, EC = 2DE, AG = 2GB
, and 
, and
So
, and 
Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal
Since
So
Hence
(ii) Since we know from above that
. So
Hence
(iii) Since height of triangle EFC and triangle EBF are equal. So
Hence
(iv) Consider the trapezium in which
(From (iii))Now from (ii) part we have
(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x
Now consider the tow triangles CFI and CBH which are similar triangles
So by the property of similar triangle CI = k and IH = 2k
Now consider the triangle EGF in which
Now
(Multiply both sides by 2)
…… (2)
From (1) and (2) we have