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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? –  BlueRaja - Danny Pflughoeft Jul 27 '10 at 19:07
    
Any angle formed by a chord with one mutual endpoint and the other on either corner of the semicircle. –  Justin L. Jul 27 '10 at 19:26
    
@Casebash: Please edit to emphasize any angle, or more specifically the angle between the two endpoints and any point on the circle. –  Tom Stephens Jul 27 '10 at 19:46
    
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The usual terminology in high school texts in the U.S. is that the angle is "inscribed" in a semicircle. –  Isaac Jul 27 '10 at 21:20
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3 Answers

up vote 13 down vote accepted

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

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

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 line –  Casebash Jul 27 '10 at 10:10
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Nice! Hadn't seen this proof before. –  Ben Alpert Jul 27 '10 at 12:46
    
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. –  T.. Aug 4 '10 at 5:48
    
@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). –  Isaac Aug 4 '10 at 5:55
    
@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. –  Isaac Aug 4 '10 at 6:11
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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. :) –  J. M. Aug 6 '10 at 14:24
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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.

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