Proof of Angle in a Semi-Circle is of $90$ degrees 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).
 A: 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.
A: In fact it is a corollary of the theorem stating that the angle subtended at the centre of the circle (here 180 degrees) is double the angle subtended at the centre. If this is not satisfactory use the contrapositive instead: if angle ABC is not 90 degrees, the angle AOB cannot be 180 degrees, i.e. A,O,B are not collinear.
A: Nonstandard proof
Consider the semi-circle with endpoints A and C and center O and the inscribed angle ∠ABC (B on the semi-circle) together with the rotation image of both about O by 180°.  The image of A is C and vice versa; let B' be the image of B.  The image of a line under a 180° rotation is parallel to the original line, AB is parallel to CB' and BC is parallel to B'A, so ABCB' is a parallelogram.  BO and its image must be parallel, but the image of O is itself, since it is the center of rotation, and if BO and B'O are parallel and contain a point in common, they must lie on the same line, so BB' passes through O.  AC and BB' (the diagonals of ABCB') are both diameters of the circle, so they are congruent.  A parallelogram with congruent diagonals is a rectangle.  Thus, ∠ABC is a right angle (and has measure 90°).
diagram http://www.imgftw.net/img/762828246.png
Standard proof (or, at least, my guess at it based on the description in the question)
As above, consider the semi-circle with endpoints A and C and center O and the inscribed angle ∠ABC (B on the semi-circle).  Draw in radius OB.  OA = OB, so △AOB is isosceles and ∠OAB≅∠OBA.  OB = OC, so △BOC is isosceles and ∠OBC≅∠OCB.  Let α=m∠OAB=m∠OBA and β=m∠OBC=m∠OCB.  In 
△ABC, the measures of the angles are α, α+β, and β, so α+(α+β)+β=180° or 2(α+β)=180° or α+β=90°, so ∠ABC has measure 90° and is a right angle.
diagram http://www.imgftw.net/img/319527897.png
edit: Another Nonstandard proof
Use the labeling as above and apply Stewart's Theorem to △ABC: $$(AB)^2(OC) + (BC)^2(AO) = (AC)((BO)^2 + (AO)(OC))$$  Substituting the length r of the radius of the semicircle as appropriate: $$(AB)^2r + (BC)^2r = 2r(r^2 + r^2)=4r^3$$  Dividing both sides by r: $$(AB)^2+(BC)^2=(2r)^2=(AC)^2$$
So, by the converse of the Pythagorean Theorem, ∠ABC is a right angle.
