# Is this a valid proof that the sum of the angles in a triangle is $180^\circ$?

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 $DF = FG$. • Read more on the fifth postulate. – Shailesh Dec 11 '15 at 9:01
• @Shailesh: You couldn't have been vaguer. – Alex M. Dec 11 '15 at 9:13
• How you developed the thought that Ut=Uq im not gettin it – Archis Welankar Dec 11 '15 at 9:28
• It can be proved that the median in a right triangle is equal to half the hypotenuse. And infact I posted a proof, check out the second and third diagrams – alexgiorev Dec 11 '15 at 9:30
• How do you know that $H$ is the midpoint of $DG$? And where are you using your results from the second diagram? – gebruiker Dec 11 '15 at 9:41

Draw a triangle ABC with vertex A at top and BC as base . Then draw a parallel line EAF to BC passing through A then see we have 2 pairs of alternate angles. Then $angle EAB=angleABC$ same for other side now note that $EAB+BAC+FAC=180..$ linear pair replace EAB with ABC and FAC with ACB see you have proven it. Hope its clear