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.

Let ABC be the triangle and D be the mid-point of AB and DE || BC.

By theorem 6.1,

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

So, In △ ABC,

$\therefore \dfrac{AD}{BD} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{AE}{EC} = 1 \text{ (Since, D is the mid-point of AB so, AD = BD.)}\\[1em] \Rightarrow AE = EC.$

Since, AE = EC.

∴ E bisects AC.

Hence, proved that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.