Question

In the figure $\Delta ABC$, D, E , F are the midpoints of side BC, CA and AB respectively. Show that
(i) BDEF is a parallelogram
(ii) $ar\left(\Delta DEF\right)=\frac{1}{4}ar\left(\Delta ABC\right)$
(ii) $ar\left(BDEF\right)=\frac{1}{2}ar\left(\Delta ABC\right)$

Answer 1

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)