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.

Let ABC be the triangle, D be the mid-point AB and E be the mid-point of AC.

So, in △ ABC,

$\dfrac{AD}{BD} = 1$ ………(1)

$\dfrac{AE}{EC} = 1$ ………(2)

From (1) and (2), we get :

$\therefore \dfrac{AD}{BD} = \dfrac{AE}{EC}$.

By 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.

Since, In △ ABC,

$\dfrac{AD}{BD} = \dfrac{AE}{EC}$.

∴ DE || BC.

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