D, E and F are a respectively the mid-points of the sides BC,
CA and AB of a triangle ABC.
Then show that:
1. BDEF is a parallelogram.
2. area (DEF 1upon2 ar (ABC).
3. area (BDFF) =1upon2 area (ABC).
Answers
Parallelogram :A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.
In a parallelogram diagonal divides it into two triangles of equal areas.
Mid point theorem:The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
SOLUTION :
Given:ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F.
To show:(i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4ar (ABC) (iii) ar (BDEF) =1/2 ar (ABC)
Proof: i)Since E and F are the midpoints of AC and AB.
BC||FE & FE= ½ BC= BD
(By mid point theorem)
BD || FE & BD= FE
Similarly, BF||DE & BF= DE
Hence, BDEF is a parallelogram
.[A pair of opposite sides are equal and parallel]
(ii) Similarly, we can prove that FDCE & AFDE are also parallelograms.
Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.
∴ ar(ΔBDF) = ar(ΔDEF) — (i)
In Parallelogram AFDE
ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)
In Parallelogram FDCE
ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)
From (i), (ii) and (iii)
ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)
ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)
4 ar(ΔDEF) = ar(ΔABC)(From eq iv)
ar(∆DEF) = 1/4 ar(∆ABC)........(v)
(iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)
ar(parallelogram BDEF) = 2× ar(ΔDEF)(From eq iv)
ar(parallelogram BDEF) = 2× 1/4
ar(ΔABC)(From eq v)
ar(parallelogram BDEF) = 1/2 ar(ΔABC)
hope it helps u..