Using Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

(N/A) Consider a triangle $ABC$ in which $P$ is the mid-point of side $AB$ and a line $PQ$ is drawn parallel to $BC$ such that it intersects $AC$ at $Q$.
According to the Basic Proportionality Theorem (Theorem $6.1$),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Substituting this into the equation,we get:
$1 = \frac{AQ}{QC}$
$\Rightarrow AQ = QC$
This proves that $Q$ is the mid-point of $AC$,meaning the line $PQ$ bisects the third side $AC$.