Lesson 7-5: The SSA Condition and HL Congruence (Day 44)

Lesson 7-5 of the U of Chicago text is on SSA and HL. I've already mentioned how I'll be able to prove HL without using AAS, since we have to wait before I can give an AAS proof.

Meanwhile, we know that SSA is invalid, but the U of Chicago text provides us with an SsA Congruence Theorem, where the size of the S's implies that it must be the longer of the congruent sides that is opposite the congruent angle. The text doesn't provide a proof of SsA because the proof is quite difficult -- certainly too difficult for high school Geometry students.

But you know how I am on this blog. I'm still curious as to what a proof of SsA entails, even if we don't ask high school students to prove it. Last year, I mentioned how SsA leads to the ambiguous case of the Law of Sines.

I've noticed that when using the Law of Sines to solve "both" triangles for the SsA case, the "second" triangle ends up having an angle sum of greater than 180 degrees. We can use the Unequal Sides Theorem to see why this always occurs -- that theorem tells us that the angle opposite the "s" must be smaller than the angle opposite the "S" (the known angle). We know that if two distinct angles between 0 and 180 have the same sine, then they are supplementary. So the second triangle would have angle sum of at least 180 minus the smaller angle plus the larger angle -- which always must be greater than 180. (This is often called the Saccheri-Legendre Theorem, named for two mathematicians with whom we're already familiar and associate with non-Euclidean geometry. Of course spherical geometry is not neutral, and Saccheri-Legendre fails in spherical geometry.)

That the sum of the angles of a triangle can never be greater than 180 (but we don't necessarily know that it's exactly 180) is neutral, but the Law of Sines is not neutral. Nonetheless, it is known that SsA is a neutral theorem. So the Law of Sines can't be behind the secret proof that U of Chicago text doesn't print in its text.

But one of the questions from the U of Chicago text hints at how to prove SsA -- and it's one that I included on my HL/SsA worksheet last year. Here is the question:

Follow the steps to make a single drawing of a triangle given the SSA condition:
a. Draw a ray XY.
b. Draw angle ZXY with measure 50 and XZ = 11 cm.
c. Draw circle Z with radius 9 cm. Let W be a point where circle Z and ray XY intersect.
d. Consider triangle XZW. Will everyone else's triangle be congruent to yours?

The answer is that they most likely won't. In step (c) we are to let W be a point -- not the point -- where the circle and the ray intersect. This implies that there could be more than one point where they intersect -- and in fact, there are two such points. But there can never be a third such point, because a circle and a ray (or line) intersect in at most two points. We can prove this indirectly using the Converse of the Perpendicular Bisector Theorem:

Lemma:
A line and a circle intersect in at most two points.

Indirect Proof:
Assume towards a contradiction that there exists a circle O that intersects line l in at least three points A, B, and C, and without loss of generality, let's say that B is between A and C. By the definition of circle, O is equidistant from A, B, and C. From the Converse of the Perpendicular Bisector Theorem, since O is equidistant from A and B, O lies on m, the perpendicular bisector of AB -- and again using the converse, since O is equidistant from B and C, O lies on n, the perpendicular bisector of BC. We know that m and n are distinct lines because m intersects l at the midpoint of AB and n intersects l at the midpoint of BC -- and those midpoints are distinct because B lies between them.

Now both m and n are said to be perpendicular to l, since each is the perpendicular bisector of a segment of l. So by the Two Perpendiculars Theorem, m and n must be parallel -- and yet O is known to lie on both lines, a blatant contradiction. Therefore a line and a circle can't intersect in three points, so the most number of points of intersection is two. QED

So let's prove SsA now using this lemma and the construction from the problem above. Once again, the proof must be indirect. Even though most of our congruence theorem proofs call the two triangles ABC and DEF, I will continue to use the letters XZW so that it matches the above question.

Given: AB = XZ < BC = ZW, Angle A = Angle X
Prove: Triangles ABC and XZW are congruent.

Proof:
We begin by performing the usual isometry that maps AB to XZ. As usual, we wish to show that the final reflection over line XZ must map C to W -- that is, C' must be W.

So assume towards a contradiction that C' is not W. As usual, the given pair of congruent angles allows us to use the Flip-Flop Theorem to map ray AC to ray XY. Just as in the problem from the text, we know that C' must be a point on ray XY (since ray AC maps to ray XY), and it must be the correct distance from Z (since reflections preserve distance). Now the set of all points that are the correct distance from Z is the circle mentioned in the above problem. We know that the intersection of a circle and a ray is at most two points, and so the assumption that C' is not W implies that W must be one of the two points of intersection, and C' must be the other point.

We must show that this leads to a contradiction in the SsA case -- that is, when AB and XZ are longer than the sides BC and ZW. We see that if C' and W are distinct points equidistant from Z, then the triangle ZC'W must be isosceles, and so its base angles ZC'W and ZWC' must be congruent.

Now we look at triangle ZC'X. It contains an angle, ZC'X, which forms a linear pair with ZC'W, so its measure must be 180 - m/ZC'W -- that is, m/ZC'X = 180 - m/ZWC' by substitution. It also contains an angle ZXC' -- which (renamed as ZXW), we see must be larger than ZWC' (renamed as ZWX) by the Unequal Sides Theorem -- ZXW is opposite the longer side ZW (in triangle ZXW), so it must be the bigger angle.

So now we add up the measures of two of the three angles in triangle ZC'W -- the angle ZC'X has measure 180 - m/ZWC', and angle ZXC' is known to be greater than ZWC'. So the sum of the two angles is greater than 180 - m/ZWC' + m/ZWC' -- that is, it is greater than 180. That is, the sum of the angles of triangle ZC'W is greater than 180 -- which is a contradiction, since by the Triangle-Sum (actually Saccheri-Legendre) Theorem, the sum must be at most 180.

Therefore the assumption that C' and W are different points is false. So C' must be exactly W. QED

Of course we wouldn't want to torture our students with this proof. It depends on three theorems -- Isosceles Triangle, Unequal Sides, and Triangle-Sum -- that we have yet to prove. On my worksheet, I added an extra note to this problem explain how it leads to SsA.