Question

D, E, F are respectively the mid points of the sides BC, CA and AB of a $\Delta ABC$. Show that
(i) BDEF is a parallelogram
(ii) $ar\left( \Delta DEF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right)$
(iii) $ar\left( BDEF \right)=\dfrac{1}{2}ar\left( \Delta ABC \right)$

Answer 1

Hint: We use the theorem of midpoint of triangles for parallel form. We also use the diagonal property of a parallelogram. We break the whole area into four smaller triangles. Then we use the theorems to prove the postulates.
D, E, F are respectively the midpoints of the sides BC, CA, and AB of a $\Delta ABC$.
So, $AF=FB,BD=DC,CE=AE$.

Complete step-by-step solution:
We know that the joining line segment of the midpoints of any two sides of a triangle is parallel to the third side of that triangle.
Also, the diagonals of a parallelogram divide the parallelogram into two triangles of equal area.

In this problem F, E are the midpoints of the sides AB and AC respectively.
So, $FE||BC$.
Similarly, $ED||AB$ and $FD||AC$ as D, E, F are respectively the midpoints of the sides BC, CA, and AB of a $\Delta ABC$.
For quadrilateral BDEF, $FE||BC$ gives $FE||BD$ as B, D, C are collinear and $ED||AB$ gives $ED||FB$ as B, F, A are collinear.
So, BDEF is a parallelogram as the opposite sides are parallel to each other.
Now, DF is a diagonal of parallelogram BDEF which gives
$ar\left( \Delta BDF \right)=ar\left( \Delta DEF \right)=\dfrac{1}{2}ar\left( BDEF \right)........(i)$
Again, for quadrilateral CDFE, $FE||BC$ gives $FE||DC$ as B, D, C are collinear and $FD||AC$ gives $FD||EC$ as C, E, A are collinear.
So, CDFE is a parallelogram as the opposite sides are parallel to each other.
Now, DE is a diagonal of parallelogram BDEF which gives
$ar\left( \Delta CDE \right)=ar\left( \Delta DEF \right)=\dfrac{1}{2}ar\left( CDFE \right)........(ii)$
For quadrilateral AFDE, $ED||AB$ gives $ED||AF$ as B, F, A are collinear and $FD||AC$ gives $FD||AE$ as A, E, C are collinear.
So, AFDE is a parallelogram as the opposite sides are parallel to each other.
Now, EF is a diagonal of parallelogram BDEF which gives
$ar\left( \Delta AFE \right)=ar\left( \Delta DEF \right)=\dfrac{1}{2}ar\left( AFDE \right)........(iii)$
From the three equation we got all the four smaller triangles are of equal area as
$ar\left( \Delta DEF \right)=ar\left( \Delta AFE \right)=ar\left( \Delta CDE \right)=ar\left( \Delta BDF \right)$
They in total make the whole area of $\Delta ABC$.
So, $ar\left( \Delta DEF \right)=ar\left( \Delta AFE \right)=ar\left( \Delta CDE \right)=ar\left( \Delta BDF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right)$.
Thus proved $ar\left( \Delta DEF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right)$.
Now, $ar\left( BDEF \right)=ar\left( \Delta DEF \right)+ar\left( \Delta BDF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right)+\dfrac{1}{4}ar\left( \Delta ABC \right)=\dfrac{1}{2}ar\left( \Delta ABC \right)$.
Thus proved.

Note: We need to remember that the area here is divided by only the diagonals. We don’t need to use the similarity of the triangles for this matter. The parallelogram also has been proved by just showing the parallel sides. We don’t need the equality of the sides.