In ΔABC, if a triangle is formed by joining the midpoints of the sides, the area of the triangle is
14
Here, D, E and F are midpoints of sides AB, AC and BC respectively.
⇒ AD = DB, BF = FC and AE = EC.
We know that the line joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side (midpoint theorem).
So, BC = 2DE
AC = 2DF
AB = 2EF
In △ABC and △DEF,
ABEF=ACDF=BCDE=21
∴△ABC∼△FED (by SSS similarity criterion)
Now, ar(△ABC)ar(△FED)=AB2EF2=2212
⇒ar(△ABC)=4×ar(△FED)
⇒ar(△FED)=14×ar(△ABC)