Question

The line-segment joining the mid-points of any two sides of a triangle

• A. Is parallel to the third side but it's dimension is not determinable
• B. Is half as long as the third side but can't say if it is parallel to it
• C. Is parallel to and half as long as the third side
• D. None of the above

Answer 1

The correct option is CIs parallel to and half as long as the third side

According mid-point theorem,

The line-segment joining the mid points of any two sides of a triangle is parallel to and half as long as the third side.

Given $ABC$ is a triangle, E and F are midpoints of the sides $AB$ , $AC$ respectively.
To prove : $EF||BC$ and EF = $\frac{1}{2}$ BC.

Construction : Draw a line $CD$ parallel to $AB$ ,it intersects $EF$ at $D$.

Proof :
In a $△AEF$ and $△CDF$

$\angle EAF=\angle FCD$ ( Alternate interior angles)

$AF=FC$ ( F is the midpoint)

$\angle AFE=\angle CFD$ ( vertically opp. Angles)

$△AEF\cong △CDF$ (ASA congruence property)

So that $EF=DF$ and $AE=CD$ ( By CPCT )

$BE=AE=CD$

$\therefore$ $BCDE$ is parallelogram.

$\Rightarrow ED\parallel BC$ (Opposite sides of parallelogram are parallel)

$\Rightarrow EF\parallel BC$

$\therefore$ $EF=DF$ (Proved)

$\Rightarrow$ $EF+DF=ED=BC$ (Opposite sides of the parallelogram are equal)

$\Rightarrow$ $EF+EF=BC$

$\therefore EF=\frac{1}{2}BC$