Using Theorem $6.2$,prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

(N/A) Consider a triangle $ABC$ in which $P$ and $Q$ are the mid-points of sides $AB$ and $AC$ respectively.
Given: $AP = PB$ and $AQ = QC$.
To prove: $PQ \parallel BC$.
Proof:
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Since $Q$ is the mid-point of $AC$,we have $AQ = QC$,which implies $\frac{AQ}{QC} = 1$.
From the above two equations,we get $\frac{AP}{PB} = \frac{AQ}{QC}$.
According to the converse of the Basic Proportionality Theorem (Theorem $6.2$),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$PQ \parallel BC$.