SSA and the Ambiguous Case in a Congruence Proof
Mr. Okafor is reviewing a two-column proof from a unit on triangle congruence. The given information reads: in triangles ABC and DEF, AB = DE = 9, BC = EF = 7, and angle A = angle D = 40°, where angle A is opposite side BC and angle D is opposite side EF. The student's final line reads "ΔABC ≅ ΔDEF by SSA" and the rest of the proof is clean. The diagram in the textbook happens to show an obtuse-looking triangle at C, which is why the student's answer looks plausible. Mr. Okafor wants to use this proof as a teaching moment about the limits of congruence criteria, and he has compasses and straightedges available for the next class. He also needs to be sure the conclusion itself is or is not defensible before he responds.
1. Determine whether the student's conclusion is justified. Support your answer with the actual angle measures implied by the given data.
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The conclusion is not justified: this is the ambiguous case of SSA, and two non-congruent triangles satisfy the data.
Use the Law of Sines in triangle ABC with $a = BC = 7$, $c = AB = 9$, and $A = 40°$: $$\sin C = \frac{c \sin A}{a} = \frac{9 \sin 40°}{7} \approx \frac{9(0.6428)}{7} \approx 0.8264.$$ So $C \approx 55.7°$ or $C \approx 124.3°$.
Check both: if $C = 55.7°$, then $B = 180° - 40° - 55.7° = 84.3°$. If $C = 124.3°$, then $B = 180° - 40° - 124.3° = 15.7°$. Both are valid triangles. Since $BC = 7 < AB = 9$ and the given angle is acute, the side opposite the given angle is too short to determine the triangle uniquely. SSA is not a congruence criterion, so the proof's final line fails even though the given numbers are consistent.
2. Mr. Okafor plans a compass-and-straightedge construction. Describe the construction that makes the ambiguity visible and explain what students should observe.
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Students should build the triangle from the angle outward, which is exactly the order SSA specifies.
- Draw a ray from A and mark off angle A = 40° to form a second ray.
- On one ray, mark B so that AB = 9.
- Set the compass to 7 and swing an arc from B. Because 7 is less than 9, the arc crosses the second ray at two distinct points, $C_1$ and $C_2$.
- Measure angle C in each triangle.
Students should observe that $C_1$ and $C_2$ give the two triangles computed earlier: the acute-angle version ($C \approx 55.7°$) and the obtuse-angle version ($C \approx 124.3°$). Both satisfy AB = 9, BC = 7, and angle A = 40°, yet the triangles are clearly different shapes. The key follow-up question is when the arc crosses only once - when $BC = AB\sin A$ (tangent, one right triangle) or when $BC \ge AB$, which is why SSA sometimes produces a unique triangle but is still not a valid general criterion.
3. A student asks, "If SSA can sometimes give one triangle, why can't we just use it when it works?" Give the response Mr. Okafor should give, and connect it to why ASA and SAS are accepted.
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The answer is about what a proof is allowed to assume.
"A congruence criterion is a promise that holds for every input. SSA does not make that promise, so a proof that cites it is invalid even when the particular numbers happen to give one triangle. To use SSA you would first have to prove that the ambiguous case cannot occur here - essentially doing extra work to show you are in a special situation. That is allowed but it is not a shortcut."
ASA and SAS are accepted precisely because no such side condition exists: once two angles and the included side are fixed, the third angle is forced and the remaining sides are determined; once two sides and the included angle are fixed, the Law of Cosines fixes the third side. The included angle is what removes freedom. A useful contrast for students: SAS locks the triangle, SSA leaves a hinge that can swing to a second position.