Question

abc is a triangle with d,e and f as midpoints of sides bc,ac and ab respectively. a line through c is drawn parallel to de meeting fe produced at g. show that ar (dcge)=1/2 ar (abc)

Answer 1

in triangle abc, d, e and f are the midpoints
therefore by midpoint theorem,
de =  1/2 of ab and de is parallel to ab
ef = 1/2 of bc and ef parallel to bc
df = 1/2 of ac and df parallel to ac

now, in quad fecd, fe is parallel dc and fe is equal to dc (d is the mid pt)
therefore, fecd is a parallelogram.
similarly, aedf and bfed are parallelograms.

we know that the diagonal of a parallelogram divides it into two triangles of equal areas.

so,
ar afe = ar fed...1
ar dec = ar fed....2
ar bfd = ar fed....3

from 1,2 and 3,

ar fed = ar afe = ar dec = ar bfd....4

now, ar dec + ar fed + ar afe + ar bfd = ar abc

from 4

4(ar dec)= ar abc  or  ar dec = 1/4 ar abc....5


now, in quad degc, de is parallel to cg (by cons) and eg is parallel to dc (proved above)

therefore, degc is a parallelogram.
 now,

ar dec + ar egc = ar degc
ar dec = ar egc ( diagonal of a parallelogram divides it into two triangles of equal areas)

so, 2(ar dec) = ar degc.......6

from 5 and 6

ar degc= 2 (1/4 of abc)

therefore, ar degc = 1/2 ar abc

hence, proved.