I’m getting ready for a geometry test and I keep second-guessing myself on congruent triangles, especially SAS vs SSA.
Here’s the setup: I’ve got two triangles, ABC and DEF. In the picture, AB = DE = 5, AC = DF = 7, and ∠A = ∠D = 40°. The catch is that the triangles look like mirror images-like the 40° angle opens in opposite directions in each drawing.
My gut says they’re congruent by SAS because the two sides and the angle between them match. But then I start worrying: does the mirror-image thing matter? Am I actually using the angle that’s between the two sides I matched, or did I accidentally fall into the SSA trap without realizing it? (My brain keeps yelling “SAS!” then “No, SSA!” and now I’m confused.)
Follow-up question: when I write something like ΔABC ≅ ΔDEF, how do I pick the order so that the corresponding vertices line up correctly? Is there a quick way to check that I’m matching the included angle to the right pair of sides? And bonus: does SSA ever work (like in right triangles), or should I just avoid it unless there’s something special given?
Sorry if I’m overthinking this-just trying not to miss an easy point on the test!
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3 Responses
Your gut’s right-this is SAS: think of two sticks joined by a fixed hinge angle; AB=5 and AC=7 with ∠A=40° at their shared endpoint locks the triangle (a mirror flip is still congruent), and to order it write ΔABC ≅ ΔDEF by lining up the equal-angle vertex A↔D, then AB↔DE gives B↔E and AC↔DF gives C↔F. Example: BC = √(5²+7²−2·5·7·cos40°) ≈ 4.51 (unique), whereas if the 40° were opposite the 5 (SSA) you’d get two triangles since 7·sin40° ≈ 4.50 < 5 < 7; SSA only proves congruence in the right-triangle HL case-see https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-triangle-congruence/a/congruent-triangles.
You have SAS here. The “included angle” is the angle formed by the two sides you know. At A it’s the angle between AB and AC, and at D it’s the angle between DE and DF. Since AB = DE, AC = DF, and ∠A = ∠D, the triangles are congruent by SAS. The mirror-image look doesn’t matter; congruence allows reflections as well as rotations and translations. A quick test to avoid the SSA trap is: does the given angle touch both of the given sides? If yes, it’s SAS; if the angle is opposite one of them, it’s SSA.
For the order in ΔABC ≅ ΔDEF, match the special vertex first: A ↔ D (the equal angles). Then follow the equal sides around that vertex: AB = DE gives B ↔ E, and AC = DF gives C ↔ F. So ΔABC ≅ ΔDEF is consistent. As a check, read around each triangle in the same direction starting from the included-angle vertex; the pairs of corresponding sides you listed should line up in the same order.
About SSA: it’s generally not sufficient because of the “ambiguous case.” With an acute given angle A and sides a (opposite A) and b (adjacent), there can be 0, 1, or 2 triangles depending on a versus b sin A and b. Special cases do work: in right triangles, “hypotenuse–leg” (HL) is a valid congruence (an SSA-type situation with the right angle), and with an obtuse given angle A there is at most one triangle-only if the side opposite A is the longer one. Which version of these SSA cases have you seen in class-HL only, or the full ambiguous-case breakdown?
You’re fine: AB = DE, AC = DF, and ∠A = ∠D sits between those matched sides, so it’s SAS (mirror image doesn’t matter); write ΔABC ≅ ΔDEF with A↔D (the included vertex), then pick B↔E and C↔F so AB↔DE and AC↔DF, and remember SSA is generally ambiguous except for the right-triangle HL case.
Example: with AB = DE = 5, AC = DF = 7, and ∠A = ∠D = 40°, the third side by the Law of Cosines is sqrt(5² + 7² − 2·5·7·cos 40°) ≈ 4.6 in both triangles, so they match up to reflection.