Question

Prove that the line joining the mid-points of any two sides of a triangle is parallelogram to and half of the third side.

Answer 1

Given: $\Delta$ABC with D and E are mid points of AB and AC respectively.

Construction: Join DE and draw CF$||$AB.

To prove: DE$||$BC; DE$=\frac{1}{2}BC$

Proof: In $\Delta$EAD and $\Delta$ECF

$EC=AE$[E is mid point]

$\angle 1=\angle 2$ [VOA]

$\angle DAE=\angle ECF$ $\left[CF//AD\right]$

By AAA $\Delta EAD\cong \Delta ECF$

$DE=EF$

$CF=AD$(By CPCT)

But $AD=BD$

Hence $CF=BD$

A side is parallel as well as equal.

Hence DBCF is a parallelogram.

So, DE$||$BC

[opposite side of parallelogram]

$DE+EF=DF$

But $DF=BC$

$DE+EF=BC$

$DE+DE=BC$. $[DE=EF]$

$2.DE=BC$

$DE=\frac{1}{2}BC$

Hence proved.