Module 02 — Transformations and Congruence
Two figures are congruent when one can be stacked on the other by sliding, flipping, or turning. Those three moves keep size and shape, so corresponding sides match and corresponding angles match.
Do not trust a sketch. Pictures lie about length. Prove it with measurements, marks, or a named shortcut.
The letters in a congruence statement are a matching list. ΔABC ≅ ΔDEF means A goes with D, B with E, C with F. Switch the letters and you switch which parts you claimed.
Each triangle has three sides and three angles. You do not need all twelve pieces. Three matching pairs, in a legal pattern, lock the rest.
Hash marks on sides and arcs on angles tell you which pairs already match. Vertical angles, a shared side, a midpoint, and a bisector are how those marks get there without a ruler.
On a coordinate grid, count when a side is horizontal or vertical. Use the distance formula when it slants: square root of (Δx)² + (Δy)².
A shared side is congruent to itself. That reason is the Reflexive Property, not Symmetric.
Included means "the piece in between the other two."
SAS picture: the marked angle sits between the two marked sides.
AAA matches angles only. The triangles are the same shape, so they are similar, not necessarily the same size. You need at least one side to lock the scale.
SSA is the ambiguous case. Two sides and a non-included angle can make zero, one, or two triangles. The course will not let you name SSA as a reason.
If a quiz shows two sides and an angle that is not between them, do not write SAS. Scan for a right angle that would turn it into HL, or for a second angle that would turn it into AAS.
CPCTC means Corresponding Parts of Congruent Triangles are Congruent. First prove the triangles congruent with SSS, SAS, ASA, AAS, or HL. Then the unmarked leftover pair is congruent too, and the reason on that line is CPCTC.
Typical flow: given marks, plus Reflexive on a shared side, plus a named shortcut, then CPCTC for the piece the question actually asked about.
Example: BF bisects ∠CBA, and ∠CFB ≅ ∠AFB. Shared side BF is Reflexive. That is ASA (angle, included side, angle). Then CF ≅ AF by CPCTC.
CD is the perpendicular bisector of AB, meeting at E. Prove C is the same distance from A as from B.
C can slide anywhere on the line except that the same SAS still works. If C sits on AB, it is the midpoint, and the distances still match.
Watch for SAS vs SSA, and CPCTC vs a Property of Equality that happens to mention the same length.