It can be proved that the median in a right triangle is equal to half the hypotenuse. Consider an aritrary triangle $QRS$ , $ST\perp QR$ and finally let $U$ and $V$ be the midpoints of $QS,SR$ respectively. Then we know that $UQ=UT$,so $\angle UQT = \angle UTQ$ and for similar reasons $\angle UST = \angle UTS,\angle VST = \angle VTS$ and finally $\angle VTR = \angle VRT$. But $\angle UTQ +(\angle UTS+\angle VTS) + \angle VTR = 180^\circ=\angle UQT + (\angle UST + \angle VST) + \angle VRT = \angle UQT + \angle USV + \angle VRT$.
This is the proof that the median in a right triangle is equal to half the hypotenuse: First I will prove that the line connecting two midpoints is equal to half the opposite and is half of it.
Now on the the actual proof: let $F$ be the midpoint of the hypotenuse and let $FH\| GE$ By the parallel postulate (and the above result) we have that H is the midpoint of DG and therefore the perpendicular bisector from which it follows that $DF = FG$.