SSA Data and a Flawed SAS Claim
During a proof review in an OST Geometry class, two students present competing arguments about triangles ABC and DEF. They are given AB = DE = 7 cm, BC = EF = 5 cm, and angle A = angle D = 35 degrees. Priya writes, “AB is congruent to DE, BC is congruent to EF, and angle A is congruent to angle D, so triangle ABC is congruent to triangle DEF by SAS.” Marcus objects that the angle is not between the two named sides. The teacher asks the class to decide whether the measurements actually force the two triangles to be congruent and to identify what evidence would settle the question. A ruler-and-protractor construction is allowed, but no diagram is supplied, so students must reason from the measurements alone. The class must also compare the height from B to line AC with BC, because that comparison determines how many different triangles can fit the data. Finally, they need to recommend one additional measurement that would make a congruence criterion apply immediately.
1. Is Priya’s use of SAS valid? Explain what SAS requires and whether her stated information meets that requirement.
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Priya’s conclusion may or may not be true, but her stated reason is not valid. SAS requires two sides and the included angle-the angle formed by those same two sides. Here the two sides named are AB and BC. The angle between them is angle B, not angle A. Angle A is formed by AB and AC. Therefore the given information is an SSA pattern, not SAS. SSA is not a general congruence criterion because it can allow two different triangles with the same two sides and non-included angle. The proof needs a valid criterion such as SAS with angle B, SSS with AC = DF, or ASA/AAS with an appropriate pair of angles.
2. Compute the height from B to line AC for the given 35-degree angle and side AB = 7 cm. Use that height and BC = 5 cm to determine how many noncongruent triangles satisfy the data. Show your work.
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Drop a perpendicular from B to line AC and call its length h. In the right triangle formed with hypotenuse AB = 7, h = 7 sin 35 degrees ≈ 7(0.5736) ≈ 4.02 cm. Since BC = 5 cm is greater than h ≈ 4.02 cm, the circle centered at B with radius 5 can meet line AC in two points. Since BC = 5 is also less than AB = 7, both intersection points lie on the ray from A, so both give valid triangles. Algebraically, let AC = x. By the Law of Cosines, 25 = 49 + x² − 14x cos 35 degrees, so x² − 11.468x + 24 = 0. The two positive roots are approximately x ≈ 8.71 and x ≈ 2.75. Thus there are two noncongruent triangles with the same AB, BC, and angle A. The given measurements alone do not force congruence.
3. Recommend one additional measurement that would let the class use a congruence criterion immediately, and explain why it works. Then state whether the original two triangles must be congruent.
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One good choice is to measure AC and DF. If AC = DF, then the triangles have AB = DE, BC = EF, and AC = DF, which is SSS. Another valid choice is to measure angle B and angle E: those are the included angles between AB and BC (and between DE and EF), so equality would give SAS. Measuring only angle C would not be enough immediately, because angle C is not included between AB and BC. The original triangles are not forced to be congruent by the given SSA data; the two possible AC lengths found earlier produce genuinely different triangles. The extra measurement removes the ambiguity and permits a valid criterion.