RHS is a well known test for determining the congruency of triangles. It is easy enough to prove it works, simply use Pythagorus' theorem to reduce to SSS. I thought that it seems strange that this only works for an angle being 90 degrees - or does it? What if I tried changed the given angle to 89 degrees or 91 degrees, would it still be uniquely identified up to congruence?

  • $\begingroup$ Seeded question $\endgroup$ – Casebash Jul 27 '10 at 6:25
  • $\begingroup$ I am indeed missing the well-known (at least from German school geometry) "two sides and the angle opposite the longer of these sides" from the (English) WP page $\endgroup$ – Hagen von Eitzen Jan 21 '14 at 13:38

Supposing that we knew two triangles had one angle congruent, a side adjacent to the angle congruent, and the side opposite the angle congruent. This is sometimes referred to as SSA, which is not a congruence theorem (and I've heard it said that it is "ass-backwards"). With a little more information, it is possible to determine congruence in some instances.

As you'd said about RHS, in that case, you can use right-triangle trigonometry to determine that the unknown sides are congruent, then use SSS to establish congruence. Without the right angle, the technique for determining the length of the third side of the triangle is to use the Law of Sines to determine the measure of the unknown angle opposite the known side, use that to find the measure of the third angle, then use the Law of Cosines to determine the length of the unknown side.

Let's call one of the triangles ABC with ∠A and AB and BC being known. From the Law of Sines, $\frac{\sin A}{BC}=\frac{\sin C}{AB}$ or $\sin C=\frac{AB\cdot\sin A}{BC}$. There will be two values of C in the range 0° to 180° that satisfy this equation, unless sin C = 1. So:

  1. if sin C = 1, then C is a right angle, the triangle is uniquely determined so congruence can be established;
  2. if A ≥ 90°, C < 90°, so there is only one solution for C that makes sense in this triangle, the triangle is uniquely determined, and congruence can be established (one might call this SSobtuseA);
  3. if A < 90°, but BC ≥ AB, then B ≥ C (in a triangle, the largest/smallest side is opposite the largest/smallest angle), so the only solution for C that makes sense in this triangle is the one with C < 90° (if B ≥ C > 90°, then A + B + C > 180°), the triangle is uniquely determined, and congruence can be established (one might call this SsA, with the relative sizes of the S/s indicating the relative lengths);
  4. otherwise (when A < 90° and BC < AB), there are two possible values for C, both of which lead to triangles, so there are two possible triangles satisfying the given information, and congruence cannot be established.

So, to your specific question, if the angle were 91° (case 2), congruence would follow; if the angle were 89°, congruence may or may not follow, depending on what you can determine about the other sides.

As an aside, RHS is also commonly referred to (at least in the midwestern U.S. in contemporary high school geometry) as hypotenuse-leg or HL.

  • $\begingroup$ Very nice. I really think that they should cover this in high school trigonometry $\endgroup$ – Casebash Jul 27 '10 at 7:05
  • 2
    $\begingroup$ Depending on the text and teacher, it is sometimes covered in high school (in the U.S.) in geometry or when trigonometry is covered (advanced algebra/precalculus). Even if it's not explicitly covered as a congruence testing issue, the basic ideas of uniquely-determining information have to be covered when talking about "triangle-solving" involving the Law of Sines (often discussed as the "ambiguous case"). $\endgroup$ – Isaac Jul 27 '10 at 7:08

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