Show that two triangles are congruent if and only if corresponding sides and angles are congruent, and explain how ASA, SAS and SSS follow from the rigid-motion definition

Learning goal: Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent, and explain how each of the triangle congruence criteria ASA, SAS and SSS follows from that definition.

The statement and its two directions. "Two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent" is two claims. The forward direction: if a sequence of rigid motions carries triangle ABC onto triangle DEF with A → D, B → E, C → F, then AB = DE, BC = EF, AC = DF and the angles at A, B, C equal the angles at D, E, F. The backward direction: if those six congruences hold, then such a sequence of rigid motions exists. "If and only if" is proved by proving both.

Forward direction. Rigid motions preserve distance and angle measure, and a sequence of rigid motions is itself a rigid motion. The sequence carries segment AB onto segment DE, so AB = DE, and similarly for the other two sides; it carries the angle at A, formed by rays AB and AC, onto the angle formed by rays DE and DF, so the angles are equal, and similarly at B and C. That is the whole proof: the definition of congruence hands us the rigid motion, and the properties of rigid motions hand us the six equalities.

Backward direction, and the pattern behind every criterion. Suppose AB = DE, BC = EF, AC = DF and the three angle pairs are equal. Build the sequence in three moves. Move 1: translate by the segment from A to D, so A lands on D. Move 2: rotate about D until the image of ray AB lies along ray DE; because AB = DE, the image of B is exactly E. Move 3: if the image of C is on the opposite side of line DE from F, reflect across line DE; D and E are on that line and stay fixed. Now A is at D, B is at E, and the image of C is on the same side of line DE as F. Because the angle at A equals the angle at D, the image of ray AC coincides with ray DF; because AC = DF, the image of C is the point of that ray at distance DF from D, which is F. Every vertex lands, so the triangles are congruent. Read the argument again and notice how little it used: the moves only needed AB = DE, and the last step only needed the angle at A and the side AC. That is the seed of SAS.

SAS from the definition. Suppose only AB = DE, the angle at A equals the angle at D, and AC = DF (two sides and the included angle). Moves 1 to 3 need only AB = DE, and they place A on D, B on E and the image of C on the F side of line DE. The equal angles at A and D put the image of ray AC on ray DF, and AC = DF puts the image of C at F. Three facts, and the sequence exists. In coordinates: triangle with A(0, 0), B(6, 0), a right angle at A and AC = 4 has C at (0, 4); triangle with D(2, 1), E(8, 1), a right angle at D and DF = 4 has F at (2, 5). The translation by 2 right and 1 up sends A to D, B to E and C to F; the third sides are forced to be equal, BC = √(36 + 16) = √52 and EF = √(36 + 16) = √52, without ever being given.

ASA from the definition. Suppose AB = DE, the angle at A equals the angle at D, and the angle at B equals the angle at E (two angles and the included side). The same three moves place A on D, B on E and the image of C on the F side. The equal angles at A and D put the image of ray AC on ray DF; the equal angles at B and E put the image of ray BC on ray EF. The image of C lies on both rays, and F lies on both rays. Two distinct rays that are not on the same line meet in at most one point, so the image of C is F. Numerically: A(0, 0), B(8, 0) with 45° angles at both A and B force C = (4, 4), and D(1, 2), E(9, 2) with 45° angles at both force F = (5, 6); the translation 1 right and 2 up carries A, B, C to D, E, F.

SSS from the definition. Suppose AB = DE, BC = EF and AC = DF. Moves 1 to 3 place A on D, B on E and the image of C on the F side of line DE. The image of C is at distance AC = DF from D, so it lies on the circle about D with radius DF; it is at distance BC = EF from E, so it lies on the circle about E with radius EF. Two circles with different centers meet in at most two points, and those two points are reflections of each other across the line of centers, line DE, so they lie on opposite sides of it. F is one of the two points; the image of C is on the same side as F; so the image of C is F. Numerically: A(0, 0), B(6, 0), AC = 5, BC = 5 gives C on the vertical line x = 3 with 9 + y² = 25, so C = (3, 4) or (3, −4); D(4, 3), E(10, 3), DF = 5, EF = 5 gives F = (7, 7) or (7, −1). Choosing C(3, 4) and F(7, 7), the translation 4 right and 3 up carries A, B, C to D, E, F.

Why SSA and AAA are not criteria. Two sides and a non-included angle can be matched by two different triangles. With A(0, 0), B(10, 0), the angle at A having tangent 3/4 (about 36.9°) and BC = 6.5, the third vertex can be C = (8.4, 6.3) or C = (4.4, 3.3): both are on the ray from A, and both are 6.5 from B, since (8.4 − 10)² + 6.3² = 2.56 + 39.69 = 42.25 = 6.5² and (4.4 − 10)² + 3.3² = 31.36 + 10.89 = 42.25. The two triangles have AC = 10.5 and AC = 5.5, so they are not congruent, and SSA cannot force the third vertex. Three angles fail for a different reason: the triangles (0, 0), (3, 0), (0, 4) and (0, 0), (6, 0), (0, 8) have the same three angles (90°, about 53.1° and about 36.9°) but sides 3, 4, 5 against 6, 8, 10, so no rigid motion can relate them.

The misconception. Students treat SAS, ASA and SSS as facts to memorize and then wonder why SSA is missing. They are not separate facts; each one is the observation that the three-move sequence still lands C on F with fewer than six given congruences, and SSA is missing because with two sides and a non-included angle the sequence can leave the image of C at a second point. Every criterion is a shortcut through the definition, and knowing the route is what makes the shortcut trustworthy.

Worked, the forward direction with numbers: Triangle ABC has A(0, 0), B(5, 0), C(2, 4); triangle DEF has D(9, 1), E(9, 6), F(5, 3). The rotation of 90° counterclockwise about the origin, (x, y) → (−y, x), followed by the translation 9 right and 1 up sends A → (0, 0) → (9, 1) = D, B → (0, 5) → (9, 6) = E, C → (−4, 2) → (5, 3) = F, so the triangles are congruent. Corresponding parts: AB = 5 and DE = 5; AC = √(4 + 16) = √20 and DF = √(16 + 4) = √20; BC = √(9 + 16) = 5 and EF = √(16 + 9) = 5. Angle at A: cosine (5 × 2 + 0 × 4) ÷ (5 × √20) = 10 ÷ 22.36 ≈ 0.447, about 63.4°; angle at D between (0, 5) and (−4, 2): cosine 10 ÷ (5 × √20), also about 63.4°. Angle at B: about 53.1°; at E, about 53.1°. Angle at C and at F: 180° − 63.4° − 53.1° ≈ 63.4°. All six pairs agree, as the forward direction says they must.

Worked, SAS in the three-move form: Given AB = DE = 6, angle A = angle D = 90°, AC = DF = 4, with A(0, 0), B(6, 0), C(0, 4) and D(2, 1), E(8, 1), F(2, 5). Move 1: translate by the segment from A to D, 2 right and 1 up: A → (2, 1) = D, B → (8, 1) = E already, C → (2, 5). Move 2: the image of ray AB already lies along ray DE, so the rotation is through 0°. Move 3: the image of C, (2, 5), is on the same side of line DE (the line y = 1) as F, so no reflection. The image of C is the point on ray DF at distance 4 from D, which is (2, 5) = F. Done with three given facts; BC = EF = √52 was never needed.

Worked, ASA as an intersection of rays: A(0, 0), B(8, 0), angles of 45° at A and at B. Ray AC leaves A at 45°, so it is the line y = x; ray BC leaves B at 45° toward the interior, so it is the line y = 8 − x. They meet where x = 8 − x, at (4, 4) = C. For D(1, 2), E(9, 2) with 45° angles, the rays are y = x + 1 and y = 11 − x, meeting at (5, 6) = F. After the translation 1 right and 2 up, A, B land on D, E, ray AC lands on ray DF and ray BC on ray EF; the rays cross once, so the image of C must be F. Check: (4 + 1, 4 + 2) = (5, 6).

Worked, SSS as an intersection of circles: A(0, 0), B(6, 0), AC = 5, BC = 5. C is on the circle x² + y² = 25 and on the circle (x − 6)² + y² = 25; subtracting gives 12x − 36 = 0, x = 3, then y² = 16, y = ±4. The two candidates (3, 4) and (3, −4) are reflections of each other across line AB. For D(4, 3), E(10, 3), DF = EF = 5 the candidates are (7, 7) and (7, −1). With C(3, 4) and F(7, 7): translate 4 right and 3 up, A → D, B → E, C → (7, 7) = F. Had F been (7, −1), a reflection across line DE (y = 3) would first send (7, 7) to (7, −1), which is move 3 doing its job.

Recording task: Draw triangle ABC with A(0, 0), B(5, 0), C(2, 4) and triangle DEF with D(9, 1), E(9, 6), F(5, 3) on one grid. Record a six-row table of corresponding parts, three sides and three angles, with the measurement of each part in both triangles (sides 5, √20, 5; angles about 63.4°, 53.1°, 63.4°), and below it the two-step sequence, rotation of 90° counterclockwise about the origin then translation 9 right and 1 up, verified on all three vertices. Then write the three-move argument (translate A to D, rotate about D, reflect across DE if needed) once in full, and mark in the margin which of the six given facts each move actually used.

Independent performance: Write an explanation, in your own words and with a diagram, of why each of SAS, ASA and SSS guarantees congruence, starting each explanation from the definition of congruence in terms of rigid motions and using the three-move sequence. Then construct, with coordinates, a pair of triangles that share two sides and a non-included angle but are not congruent. Success criteria: each explanation names the three moves and states which given fact is used at each step; the ASA explanation says why two rays meet in at most one point; the SSS explanation says why two circles meet in at most two points and why those points are on opposite sides of the line of centers; the SSA counterexample gives two sets of coordinates with the shared measurements computed; and a closing paragraph explains why a triangle cannot be pinned down by its three angles alone.