On Triangle Congruence, and Why SSA Does Not Work

Those who have taught geometry, when teaching triangle congruence, go through a familiar pattern. SSS (side-side-side) triangle congruence is usually taught first, as a postulate, or axiom — a statement so obvious that it requires no proof (although demonstrations certainly do help students understand such statements, even if rigorous proof is not possible). Next, SAS (side-angle-side) and ASA (angle-side-angle) congruence are taught, and most textbooks also present them as postulates. AAS (angle-angle-side) congruence is different, however, for it need not be presented without proof, for it follows logically from ASA congruence, paired with the Triangle Sum Theorem. With such a proof, of course, AAS can be called a theorem — and one of the goals of geometricians is to keep the number of postulates as low as possible, for we dislike asking people to simply accept something, without proof.

At about this point in a geometry course, because the subject usually is taught to teenagers, some student, to an audience of giggling and/or snickering, will usually ask something like, “When are we going to learn about angle-side-side?”

The simple answer, of course, is that there’s no such thing, but there’s a much better reason for this than simple avoidance of an acronym which many teenagers, being teenagers, find amusing. When I’ve been asked this question (and, yes, it has come up, every time I have taught geometry), I accept it as a valid question — since, after all, it is — and then proceed to answer it. The first step is to announce that, for the sake of decorum, we’ll call it SSA (side-side-angle), rather than using a synonym for a donkey (in all caps, no less), by spelling the acronym in the other direction. Having set aside the silliness, we can then tackle the actual, valid question: why does SSA not work?

This actually is a question worth spending class time on, for it goes to the heart of what conjectures, theorems, proof, and disproof by counterexample actually mean. When I deal with SSA in class, I refer to it, first, as a conjecture:  that two triangles can be shown to be congruent if they each contain two pairs of corresponding, congruent sides, and a pair of corresponding and congruent angles which are not included between the congruent sides, of either triangle. To turn a conjecture into a theorem requires rigorous proof, but, if a conjecture is false, only one counterexample is needed to disprove its validity. Having explained that, I provide this counterexample, to show why SSA does not work:

no SSA

In this figure, A is at the center of the green circle. Since segments AB and AC are radii of the same circle, those two segments must be congruent to each other. Also, since congruence of segments is reflexive, segment AD must be congruent to itself — and, finally, because angle congruence is also reflexive, angle D must also be congruent to itself.

That’s two pairs of corresponding and congruent segments, plus a non-included pair of congruent and corresponding angles, in triangle ABD, as well as triangle ACD. If SSA congruence worked, therefore, we could use it to prove that triangle ABD and triangle ACD are congruent, when, clearly, they are not. Triangle ACD contains all the points inside triangle ABD, plus others found in isosceles triangle ABC, so triangles ABD and ACD are thereby shown to have different sizes — and, by this point, it has already been explained that two triangles are congruent if, and only if, they have the same size and shape. This single counterexample proves that SSA does not work.

Now, can this figure be modified, to produce an argument for a different type of triangle congruence? Yes, it can. All that is needed is to add the altitude to the base of isosceles triangle ABC, and name the foot of that altitude point E, thereby creating right triangle AED.


It turns out that, for right triangles only, SSA actually does work! The relevant parts of the right triangle, shown in red, are segment DA (congruent to itself, in any figure set up this way), segment AE (also congruent to itself), and the right angle AED (since all right angles are congruent to each other). However, as I’ve explained to students many times, we don’t call this SSA congruence, since SSA only works for right triangles. To call this form of triangle congruence SSA (forwards or backwards), when it only works for some triangles, would be confusing. We use, instead, terms that are specific to right triangles — and that’s how I introduce HL (hypotenuse-leg) congruence, which is what SSA congruence for right triangles is called, in order to avoid confusion. Only right triangles, of course, contain a hypotenuse.

This is simply one example of how to use a potentially-disruptive student question — also known as a teenager being silly — and turn it around, using it as an opportunity to teach something. Many other examples exist, of course, in multiple fields of learning.

The 9-81-90 Triangle

In a previous post (right here), I explained the 18-72-90 triangle, derived from the regular pentagon. It looks like this:

18-72-90 triangle

I’m now going to attempt derivation of another “extra-special right triangle” by applying half-angle trigonometric identities to the 18º angle. After looking over the options, I’m choosing cot(θ/2) = csc(θ) + cot(θ). By this identity, cot(9°) = csc(18°) + cot(18°) = 1 + sqrt(5) + sqrt[2sqrt(5) + 5].

Since cotangent equals adjacent over opposite, this means that, in a 9-81-90 triangle, the side adjacent to the 9° angle has a length of 1 + sqrt(5) + sqrt[2sqrt(5) + 5], while the side opposite the 9° angle has a length of 1. All that remains, now, is to use the Pythagorean Theorem to find the length of the hypotenuse.

By the Pythagorean Theorem, and calling the hypotenuse h, we know that h² = (1)² + {[1 + sqrt(5)] + sqrt[2sqrt(5) + 5]}² = 1 + {2[(1 + √5)/2] + sqrt[(2√5) + 5]}² = 1 + {2φ + sqrt[(2√5) + 5]}², where φ = the Golden Ratio, or (1 + √5)/2, since I want to use the property of this number, later, that φ² = φ + 1.

Solving for h, h = sqrt(1 + {2φ + sqrt[(2√5) + 5]}²) = sqrt{1 + 4φ² + (2)2φsqrt[(2√5) + 5] + (2√5) + 5} = sqrt{6 + 4(φ + 1) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{6 + 4φ + 4 + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 4[(1 + √5)/2] + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 2 + (2√5) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{12 + (4√5) + 4[1 + √5)/2]sqrt[(2√5) + 5]} = sqrt{12 + (4√5) + (2 + 2√5)sqrt[(2√5) + 5]}, the length of the hypotenuse. Here, then, is the 9-81-90 triangle:

9-81-90 triangle


The 15-75-90 Triangle


The 15-75-90 Triangle

As the 30-60-90 triangle is based on an equilateral triangle, the 45-45-90 triangle is based on a square, the 18-72-90 and 36-54-90 triangles are based on the regular pentagon (see https://robertlovespi.wordpress.com/2013/03/12/the-18-72-90-and-36-54-90-triangles/), and the 22.5-67.5-90 triangle is based on the regular octagon (see previous post), so the 15-75-90 triangle is based on the regular dodecagon, shown here with three radii (red) and a single diagonal (purple). The 15-75-90 triangle is shown in yellow. An argument from symmetry is sufficient to show that angle EFC is the right triangle in this triangle, and the larger of its two acute angles (angle FCE) is one-half of an interior angle of this dodecagon. The interior angle of a regular decagon measures 150 degrees (the proof of this is trivial), and so angle FCE must measure half that amount, or 75 degrees. This leaves 15 degrees for angle CEF, via the triangle sum theorem.

What about the side lengths of the 15-75-90 triangle, though? First, consider the red diagonals shown, and let them each have a length of 2. Angles DAF and FAE each measure 30 degrees, since 360/12 = 30, and they are central angles between adjacent radii. This makes angle DAE 60 degrees by angle addition, and triangle DAE is known to be isosceles, since the two red sides are radii of the same regular dodecagon, and therefore are congruent. By the isosceles triangle theorem and triangle sum theorem, then, angles ADE and AED each also measure (180-60)/2 = 60 degrees, so triangle ADE is therefore equilateral, with the purple side, DE, also having a length of two. Symmetry is sufficient to see that DE is bisected by radius AC, which leads to the conclusion that EF, the long leg of the 15-75-90 triangle, has a length of 1.

Segment AF is a median, and therefore also an altitude, of equilateral triangle ADE, and splits it into two 30-60-90 triangles, one of which is triangle AEF. Its hypotenuse, AE, is already known to have a length of 2, while its short leg, EF, is already known to have a length of 1. Segment AF is therefore the long leg of this 30-60-90 triangle, with a length of √3.

AF, length √3, and FC, the short leg of the 15-75-90 triangle, together form dodecagon radius AC, already set at length 2. By length subtraction, then, FC, the 15-75-90 triangle’s short leg, has a length of 2 – √3. A test is prudent at this point, by taking the tangent of the 15 degree angle FEC in the yellow triangle. Tan(15 degrees) is equal to 0.26794919…, which is also the decimal approximation for FC/EF, or (2 – √3)/1.

All that remains to know the length ratios for the sides of the 15-75-90 triangle is to determine the length of EC, its hypotenuse, via the Pythagorean Theorem. The square of length EC must equal the square of 1 plus the square of (2 – √3), so EC, squared, equals 1 + 4 – 4√3 + 3, or 8 – 4√3. The hypotenuse (EC) must therefore be the square root of 8 – 4√3, which is √(8-4√3)) = 2√(2-√3)).

The short leg:long leg:hypotenuse ratio in a 15-75-90 triangle is, therefore, (2-√3):1:2√(2-√3)).

The 18-72-90 and 36-54-90 Triangles

It is well-known that an altitude splits an equilateral triangle into two 30-60-90 triangles, and that a diagonal splits a square into two 45-45-90 triangles. The properties of these “special right triangles,” as they are often called, are well-understood, and shall not be described here.

What happens if other polygons are split by diagonals, altitudes, or pieces thereof? Can more triangles be found which can allow, for example, exact determination of certain trigonometric ratios?

Yes, and the logical place to start looking is in the regular pentagon.


In this diagram, the yellow triangle is the 18-72-90 triangle. Its hypotenuse is a diagonal of the pentagon, and its short leg is a half-side of the pentagon. Since sides and diagonals of regular pentagons are in the Golden Ratio, (1 + √5)/2, these two sides must be in twice that ratio. Let their lengths, then, be 1 (short leg) and 1 + √5 (hypotenuse), for those are simple, and in the specified ratio. The Pythagorean Theorem may then be applied to find the length of the long leg; the result is sqrt((2√5) + 5). Yes, nested radicals appear at this point, and they resist efforts to make them go away. No one promised this would be simple!

The blue triangle is the 36-54-90 triangle. Its long leg is a half-diagonal of the pentagon, while its hypotenuse is a full side of the pentagon. These triangle sides must, therefore, be in half the Golden Ratio, so the simplest lengths for those sides (which work) are 1 + √5 for the long leg, and 4 for the hypotenuse. Applying the Pythagorean Theorem to find the length of the short leg, nested radicals appear again in the solution:  sqrt(10 – 2√5).