Question

if D,E and F are the midpoints of side BC,CA and AB respectively of a triangle ABC then using co-ordinate geometry prove that area of triangle DEF is equal to one by fourth of the area of triangle ABC ​

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here D,E and F are the midpoints of side BC,CA and AB respectively of a triangle ABC.

Let BDEF is parallelogram with FD as diagonal

As the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔBFD) = area(ΔDEF)     [1]

as AFDE is parallelogram with FE as diagonal

the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔAFE) = area(ΔDEF)      [2]

as CEFD is parallelogram with DE as diagonal

the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔEDC) = area(ΔDEF)      [3]

From Eq  (1), (2), (3), we get

area(ΔDEF) = area(ΔBFD) = area(ΔAFE) = area(ΔEDC)    [4]

from figure

⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔAFE) + area(ΔEDC)

Using Eq (4)

⇒ area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)

⇒ area(ΔABC) = $4 \times$ area(ΔDEF)

⇒ area(ΔDEF) =  $\frac{1}{4} \times$ area(ΔABC)

Hence proved.