I think I accidentally found a proof of the famous theorem that the sum of the angles of a triangle add up to $180 ^\circ$, but am not sure if it is correct. Here it is:
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.
From the diagram I hope you can see why $CTW\cong VWA$ and $VAZ\cong UBZ$ and $CT\| BU$ and $CT= BU$,so $CB = TU$ and $CB\| TU$. Using simple arithmetic you can see that $WZ = \frac{TU}{2}$.
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 $DF = FG$.
Your proof is correct: it lies upon the theorem that the median in a right triangle is equal to half the hypotenuse, which in turn is a consequence of Euclid's Proposition 29 (if a transversal intersects two parallel lines, then the alternate interior angles are congruent) and of its corollaries.
But the theorem on the sum of the angles of a triangle is usually regarded as more fundamental than the theorem of the median in a right triangle, that's why a proof depending directly on Euclid's 29 is preferred.