Question

19. In a treiam ABC, ABDC and DC-2A EF || AR, where E and F be on BC and A
Diagonal intersects EF G. Prove that, 7EF-11AR.​

Answer 1

ABCD is a IIgm diagonals AC and BD interest at O. E, F, G and H are the mid-points of OA, OD, OC and OB respectively.

In ΔAOD, E and F are mid points of the sides OA and OD respectively.   

                ∴ EF = 21AD⟶(1)[Mid−pointtheorem]

    Similarly,     FG = 21CD⟶(2)

                        GH = 21BC⟶(3)

                        HE = 21AB⟶(4)

Adding, (1), (2), (3) & (4) we get

           EF+FG+GH+HE=21(AD+CD+BC+AB)

⇒AD+CD+BC+BAEF+FG+GH+HE=1/2

Answer 2

Answer:

Step-by-step explanation:

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.