Consider a triangle ABC. Let D is the mid point of side BC and let EF is a line segment parallel to BC and E , F lie on side AB , AC respectively.
How can one prove that if the median AD of triangle cuts the line seqment EF at P then P is mid point of EF as well. Ie in other words joining mid points of sides parallel to BC we get the median through A.
It follows from $$\frac{EP}{BD} = \frac{AE}{AB} = \frac{AF}{AC} = \frac{FP}{CD} \ \ \ \text{and } \ \ \ BD = CD \implies EP = FP$$ Thus $P$ is the midpoint of $EF$.