For any triangle ABC, if D and E are on AB and AC respectively, then DE || BC and DE = 1/2 BC iff D, E are midpoints of AB and AC respectively.

To proof this theorem,

Comparing triangles AEF and CDF,

<EAF = <FCD (alternate interior angles as AB || CD)

AF = FC (F is the mid point of AC)

<AFE = <CFD (Vertically opposite angles)

Therefore, triangle AEF is similar to CDF (ASA)

So, EF = DF and AE = DC

Therefore, BE = AE = DC

Hence, BCDE is a parallelogram. (BE = DC and BE || DC)

This gives EF || BC.

Also, EF = DF and EF + DF = ED = BC

Therefore, 2EF = BC

--> EF = 1/2 BC.

A sample question on the midpoint theorem :

Find x.

From Pythagoras' Theorem, we get that the hypotenuse is 10.

The midpoint theorem implies that x is half of the hypotenuse.

Therefore, x = 10/2 = 5.