Question

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

Answer 1

In $△ABC$, $P$ and $Q$ are mid points of side $AB$ and $AC$, respectively.

$AP=PB$ and $AQ=QC$

$\frac{AP}{PB}=\frac{AQ}{QC}=1$

Using converse theorem of Proportionality (If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.) we get,

$PQ\parallel BC$ Hence Proved.