Question

D,E and F are respectively the mid - points of the sides BC, CA and AB of a ABC show that
1) BDEF is a parallelogram.
2) \[ar\left( \Delta DEF \right)\] = $\dfrac{1}{4}$\[ar\left( \Delta ABC \right)\]
3) \[ar\left( BDEF \right)\] = $\dfrac{1}{2}$​ \[ar\left( \Delta ABC \right)\]

Answer 1

Hint: To solve this question we should initially draw the figure from the data and then we can prove BDEF is a parallelogram. Then we can use different properties to prove the following questions. Prove each question accordingly considering the given conditions.

Complete step by step solution:
1) From this given question, D and E are mid points of sides BC and AC respectively.

So, DF \[|\text{ }\!\!|\!\!\text{ }\] BA $\Rightarrow $ DF \[|\text{ }\!\!|\!\!\text{ }\] BE
Similarly, FE \[|\text{ }\!\!|\!\!\text{ }\] BD. So, BDEF is a parallelogram.
Similarly, DCEF and AFDE are parallelograms.
Hence proved.
2) Now , DE is a diagonal of parallelogram BDEF.
Therefore,
\[ar\left( \Delta BDE \right)\] = \[ar\left( \Delta DEF \right)\] ……….(1)
DF is a diagonal of parallelogram DCEF.
So, \[ar\left( \Delta DCF \right)\] = \[ar\left( \Delta DEF \right)\] ……….(2)
FE is a diagonal of parallelogram AFDE.
\[ar\left( \Delta AEF \right)\] = \[ar\left( \Delta DEF \right)\]………….(3)
From 1,2 and 3, we have,
\[ar\left( \Delta BDE \right)\] = \[ar\left( \Delta DCF \right)\] = \[ar\left( \Delta AEF \right)\] = \[ar\left( \Delta DEF \right)\]
But, \[ar\left( \Delta BDE \right)\] + \[ar\left( \Delta DCE \right)\] + \[ar\left( \Delta AEF \right)\] + \[ar\left( \Delta DEF \right)\] = \[ar\left( \Delta ABC \right)\]
So 4 \[ar\left( \Delta DEF \right)\] = \[ar\left( \Delta ABC \right)\]
$\Rightarrow $\[ar\left( \Delta DEF \right)\] = $\dfrac{1}{4}$ \[ar\left( \Delta ABC \right)\]
Hence proved.
3) Now, \[ar~\left( |{{|}^{gm}}BDEF \right)\] = 2 \[ar\left( \Delta DEF \right)\]
From the above proof we know that \[ar\left( \Delta DEF \right)\] = $\dfrac{1}{4}$ \[ar\left( \Delta ABC \right)\]
So we can write this as,
\[ar~\left( |{{|}^{gm}}BDEF \right)\] = 2 $\times $ $\dfrac{1}{4}$ \[ar\left( \Delta ABC \right)\] = $\dfrac{1}{2}$ \[ar\left( \Delta ABC \right)\]
Hence proved.

Note: The student usually goes wrong in considering the figure from the given data. The proper usage of the area of the triangle and area of the parallelogram to prove the given problems. Students should be careful to avoid the mistakes that can occur.