TITLE: Proof of Angle in a Semi-Circle is of $90$ degrees
QUESTION [6 upvotes]: There is a well known theorem often stated as the angle in a semi-circle being $90$ degrees. To be more accurate, any triangle with one of its sides being a diameter and all vertices on the circle has its angle opposite the diameter being $90$ degrees. The standard proof uses isosceles triangles and is worth having as an answer, but there is also a much more intuitive proof as well (this proof is more complicated though).

REPLY [4 votes]: Really short vector proof:
Center the circle at the origin, and scale to have radius 1. Let the vertex of the right triangle be at vector $v$, and let the diameter be the segment from the vector $w$ to $-w$. 
Then
$(v-w) \cdot (v-(-w)) = (v-w) \cdot (v+w) = (v \cdot v) - (w \cdot w) = 1 - 1 = 0$, so the angle formed by $vw$ and $v(-w)$ is a right angle.