Question

The four triangle formed by joining the mid-points of the sides of a triangle respectively are:

• A. similar, not necessarily congruent
• B. congruent
• C. equilateral
• D. isosceles

Answer 1

Answer: (B)

congruent

In $\Delta ABC$, $D,E$ and $F$ are the mid points of sides $AB,BC$ and $AC$ respectively.

Since in $\Delta ABC$, $F$ is the midpoint of $AC$ and $D$ is the mid point of $AB$, therefore,

$FD=\frac{1}{2}CB$ and $FD\parallel CB$ ( by mid point theorem)

$\Rightarrow$$FD=CE$ and $FD\parallel CE$ ..........(1)

Similarly,

$DE=FC$ and $DE\parallel FC$.........(2)

$FE=DB$ and $FE\parallel DB$.........(3)

Therefore from the equations (1), (2) and (3), we get,

$ADEF,DBEF$ and $DECF$ are parallelograms.

The diagonal of a parallelogram divide it into two triangles which will be congruent to each other, thus,

$\Delta DEF\cong \Delta ADF$............(4)

$\Delta DEF\cong \Delta DBE$.........(5)

$\Delta DEF\cong \Delta FEC$..........(6)

From equations (4), (5) and (6), we get

$\Delta DEF\cong \Delta ECF\cong \Delta DBE\cong \Delta ADF$

Hence, option B is correct.