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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).

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    "the angle in a semi-circle" - sorry, which angle?2010-07-27
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    Any angle formed by a chord with one mutual endpoint and the other on either corner of the semicircle.2010-07-27
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    @Casebash: Please edit to emphasize *any* angle, or more specifically the angle between the two endpoints and *any* point on the circle.2010-07-27
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    http://science.jrank.org/article_images/science.jrank.org/circle-theorem.2.gif2010-07-27
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    The usual terminology in high school texts in the U.S. is that the angle is "inscribed" in a semicircle.2010-07-27
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    Does it have to be a "synthetic geometry" proof? Using analytic geometry for proving this looks like it would make for something shorter.2010-08-06
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    @J. Mangaldan: Shorter, but probably less intuitive. However, if you have anything that you think is interesting, go ahead and post it!2010-08-06
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    Yes, probably not intuitive, one merely takes the slopes of the two segments comprising the inscribed angle, and then check that they are orthogonal.2010-08-06

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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.

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    Nice, but I think it is easier to show B'B passes through the center by noting that B'O is just BO rotated 180 degrees and so is a straight line2010-07-27
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    Nice! Hadn't seen this proof before.2010-07-27
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    The nonstandard proof is not a proof, it is a reduction to an equivalent statement, ("parallelogram with congruent diagonals is a rectangle") that is then established using an argument identical to the standard proof. So you might as well give the standard proof directly.2010-08-04
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    @T..: I'm sure I can prove that a parallelogram with congruent diagonals is a rectangle without using the inscribed angle theorem or anything like it. For instance, I could prove it analytically: Let the vertices of the parallelogram be (0,0), (a,b), (a+c,b), and (c,0). Applying the distance formula to the diagonals, squaring both sides: $(a+c)^2+b^2=(a-c)^2+b^2$, so 4ac=0 and either a=0 or c=0. If c=0, then the parallelogram is degenerate, so a=0, which makes the figure a rectangle (the perpendicularity of the axes is not a consequence of the inscribed angle theorem).2010-08-04
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    @T..: Alternately, let ABCD be a parallelogram with congruent diagonals, AC≅BD. Opposite sides of a parallelogram are congruent, so AD≅BC. AB is congruent to itself. So, △ABC≅△BAD (SSS triangle congruence), and m∠BAC=m∠ABD. Consecutive angles in a parallelogram are supplementary (follows from the theorems about a pair of parallel lines cut by a transversal), so m∠BAC=180°-m∠ABD. Taken together, this yields m∠BAC=m∠ABD=90°, so ABCD is a rectangle.2010-08-04
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    Showing that you can set up a coordinate system and establishing the whole Cartesian geometry package reduces the problem to a large collection of synthetic proofs quite likely including the configuration one is trying to avoid. e.g., the analytical proof assumes that coordinate distance is calculated by Pythagoras' formula, and that a parallelogram's coordinates have one vector being the sum of two others (the third vertex being (a+c,b) instead of some random (x,y) ).2010-08-04
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    The alternate proof about rectangles again reveals the problem with the nonstandard "proof". In the alternate proof you show (using BA = AB as the key fact) that a rectangle possesses reflection symmetry. This is the same as the reflection symmetry of the isosceles triangles in the standard proof (of the angle-in-semicircle theorem). The nonstandard proof purports, impossibly, to be able to establish orientation-reversing reflection symmetry using only invariance under the orientation-preserving 180 degree rotation.2010-08-04
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    If you start with a unit segment, say from a point we'll call (0,0) to a point we'll call (1,0) and use only compass and straightedge, you can extend that segment arbitrarily and construct a perpendicular line to that segment through (0,0) to get axes. Given a point P somewhere in the first quadrant and a point Q somewhere on the horizontal axis, you can construct a line parallel to OQ through P and parallel to OP through Q. Call the intersection of these lines R. From there, it's a matter of applying triangle congruence and the Pythagorean theorem. Still no inscribed angle theorem use.2010-08-04
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    As to the alternate proof, equating the reflexive property of congruence and reflection symmetry of isosceles triangles seems like a bit of a stretch to me. I made no symmetry arguments in my original proof, nor in either of my proofs that a parallelogram with congruent diagonals is a rectangle. Ultimately, any proof comes down to what is assumed and you seem to think that I am assuming things that I am not assuming.2010-08-04
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    Your proofs are a symmetry argument; this does not require using the word "symmetry" in your proof. It involves establishing (what is equivalent to the conclusion of the theorem) that some configuration of points in the proof has a particular symmetry. Specifically, relabelling the diagram in the nonstandard proof as ABCD (D instead of B'), and following on that diagram your "alternate" synthetic proof that ABCD is a rectangle, the latter is identical in substance to the standard proof: both are demonstrating symmetry of the configuration across the bisector of angle AOB.2010-08-04
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    More: the 5-point configuration OABCD (a parallelogram with equal diagonals, or equivalently an angle in a semicircle plus the point opposite), is isometric to OBADC. Selecting any list of 3 or more points from OABCD, not all on a line, one has enough length and angle equations (these obey the symmetry) to show that pointset isometric to its partner in OBADC; that ABCD is a rectangle then follows by angle chasing. Any such pointset suffices for proving the theorem, there are many equivalent choices. Standard proof chooses A-O-B, "alternate" uses A-B-C; cosmetic rearrangement of same proof.2010-08-04
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    @T..: If I correctly understand your argument, you're claiming that any proof that the angle inscribed in a semicircle has measure 90° is essentially the same because it's proving the same thing, so any proof that doesn't look exactly like the standard proof is wrong. If this is the case, we fundamentally disagree about what constitutes a valid proof.2010-08-04
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    I explained a precise notion of equivalence as it applies to two specific proofs: they utilize the same configuration, and they construct the same symmetry of the configuration (invariance under reflection in the line bisecting AOB) in the same way: by choosing a sub-configuration and, in the key step of both proofs, using a congruence axiom to identify that isometry as the one between the chosen configuration and its partner. The same sort of angle-chasing cleanup then follows in both proofs. This is a precise matching of the logic and the constructions of those 2 proofs.2010-08-04
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    Diagrams return 404...2014-01-09
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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.

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    This is essentially the "not very intuitive" analytic geometry proof I was alluding to. :)2010-08-06
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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.