Congruent shapes are identical in shape and size — one maps exactly onto the other via reflection, rotation, or translation (never an enlargement, since that changes size). Similar shapes share the same shape but can differ in size, with sides in a constant ratio rather than equal. Congruence is really the special case of similarity where the scale factor is exactly $1$.

AAA only fixes the shape, not the size — the triangle can be "zoomed" to any scale while keeping every angle the same, so it proves similarity, not congruence.

SSA doesn't fix a unique triangle either: the non-included side can "swing" to meet the base at two different points, producing two genuinely different triangles that both satisfy the same SSA data (the same ambiguity behind the sine rule's ambiguous case).

State three matching pairs, each with its own justification, then conclude with the named condition:

1. Statement 1 + reason (e.g. "Given", "common side", "vertically opposite angles are equal").
2. Statement 2 + reason.
3. Statement 3 + reason.
4. Conclusion: "Therefore $\triangle\ldots\cong\triangle\ldots$ by [SSS/SAS/ASA/AAS/RHS]."

Use CPCTC — Corresponding Parts of Congruent Triangles are Congruent — to deduce any other equal side or angle you actually need. This is usually the real point of the question; the congruence proof is a stepping stone, not the final answer.

$ABC$ is equilateral. $D$ lies on $BC$ with $AD\perp BC$. (a) Prove $\triangle ADB\cong\triangle ADC$. (b) Hence prove $BD=DC$.

(a)

1. $\angle ADB=\angle ADC=90°$ — given, $AD\perp BC$.
2. $AB=AC$ — sides of an equilateral triangle are equal.
3. $AD$ is common to both triangles.
4. Conclusion: $\triangle ADB\cong\triangle ADC$ by RHS (right angle, hypotenuse $AB=AC$, common side $AD$).

(b) Since the triangles are congruent, CPCTC gives $BD=DC$ directly (corresponding sides opposite the equal angles $\angle BAD$ and $\angle CAD$).

Split it with a diagonal into two triangles, prove those triangles congruent, then use CPCTC.

Worked example: Prove opposite sides of a parallelogram $ABCD$ are equal. Draw diagonal $AC$.

1. $\angle BAC=\angle DCA$ (alternate angles, $AB\parallel DC$).
2. $AC$ is common.
3. $\angle BCA=\angle DAC$ (alternate angles, $AD\parallel BC$).
4. $\triangle ABC\cong\triangle CDA$ by ASA, so by CPCTC, $AB=CD$ and $AD=BC$.

• Quoting "SAS" when the angle you found isn't actually the included one.
• Using AAA (equal angles only) as if it proved congruence.
• Writing a fact without its justification.
• Circular reasoning — using the conclusion of a later part as a premise in an earlier one.