Please solve question no.3.

Dear Student

2.

Given that D, E and F are the mid points of sides AB, BC, CA respectively.

To show: ΔABC is divided into four congruent triangles

Proof: D is the mid point of AB

F is the mid point of AC.

∴DF||BC    (By mid point theorem)

⇒DF||BE    ......(1)

also E is the mid point of BC

and F is the mid point of AC.

∴EF||AB    (By mid point theorem)

⇒EF||DB    ......(2)

By (1) & (2)

BEFD is a parallelogram

⇒ΔBDEΔDEF   (Since diagonal of a parallelogram divides it into two congruent triangles)  ......(3)

Similarly

ΔDEFΔCEF    ......(4)

ΔDEFΔADF    ......(5)

By (3), (4) & (5)

We have,

Hence proved

2.

  

 

ΔABC is isosceles . Therefore, AB = AC ......(1)

D is the mid point of AB

E is the mid point of BC.

∴DE||BC and DE = AC/2   (By mid point theorem) .......(2)

F is midpoint of AC

E is the midpoint of BC

So, EF||AC and EF= AB/2   (By mid point theorem) .......(3)

$$\begin{gathered} From 2, AC=2DE \\ From 3, AB=2FE \\ As AB = AC \\ so, 2DE\hspace{0.17em}= 2FE \\ \Rightarrow DE\hspace{0.17em}= FE \\ So ∆DEF is isosceles \end{gathered}$$

Regards,