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. (Recall that you have proved it in Class IX).
A line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side.
Statement to prove: In a triangle, a line drawn through the mid-point of one side and parallel to another side bisects the third side.
Let \(\triangle ABC\) be any triangle. Let \(D\) be the mid-point of side \(AB\), so
\[AD = DB.\]
Through \(D\), draw a line parallel to \(BC\) which meets \(AC\) at \(E\). We must prove that \(E\) is the mid-point of \(AC\), i.e. \(AE = EC\).
Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those two sides in the same ratio.
Here, in \(\triangle ABC\), line \(DE\) is drawn parallel to \(BC\), meeting \(AB\) at \(D\) and \(AC\) at \(E\). So, by Theorem 6.1:
\[\frac{AD}{DB} = \frac{AE}{EC}. \quad (1)\]
Since \(D\) is the mid-point of \(AB\), we have
\[AD = DB \Rightarrow \frac{AD}{DB} = 1. \quad (2)\]
Substitute (2) into (1):
\[1 = \frac{AE}{EC}.\]
This implies
\[AE = EC.\]
Thus, point \(E\) divides side \(AC\) into two equal parts, i.e. \(E\) is the mid-point of \(AC\).
Therefore, a line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side.