What 'congruent' really claims
Two triangles are congruent when you could pick one up, slide it, turn it, and if needed flip it over, and have it land exactly on top of the other — every vertex on a vertex, every side on a side. A triangle has six parts: three sides and three angles. To say triangle ABC and triangle DEF are congruent triangles is to claim all six matchings at once: AB = DE, BC = EF, CA = FD, and m(angle A) = m(angle D), m(angle B) = m(angle E), m(angle C) = m(angle F). We write this as triangle ABC = triangle DEF, and the order of the letters is a promise about which part matches which.
Checking all six parts is honest but exhausting, and it is more than nature requires. A triangle is a remarkably rigid shape: fix the right three measurements and the other three have no freedom left. The whole business of this guide is learning which three are enough — and, just as importantly, which three are not.
SSS, SAS, ASA: the three workhorses
The first criterion is the most physical. If the three sides of one triangle equal the three sides of another, the triangles are congruent — this is SSS (side-side-side). Think of building a triangle from three rigid sticks of fixed lengths: once you join them end to end, the shape cannot wobble. A triangle braces itself; this is exactly why bridges and roof trusses are full of triangles and not squares, which sag into rhombuses. Three side lengths leave nothing to choose.
The second is SAS (side-angle-side): two sides and the angle between them. If two sides and their included angle in one triangle match those in another, the triangles are congruent. Picture two sticks hinged at a corner. The hinge sets the angle; the two stick lengths are fixed; so the gap between their far ends — the third side — is forced, and with it the whole triangle. The word 'included' is the load-bearing one: the angle must sit between the two named sides, snug in the corner they share.
The third is ASA (angle-side-angle): two angles and the side between them. If two angles and their included side match, the triangles are congruent. Here the picture is a baseline of fixed length with a ray leaving each end at a fixed angle. Those two rays must cross at exactly one point, pinning the apex and the whole triangle. There is also a close cousin, AAS, where the known side is not between the two angles; it works too, because the triangle angle-sum (angles total 180 degrees) lets you compute the third angle, quietly turning AAS back into ASA.
SSS : three sides match -> congruent SAS : two sides + the angle BETWEEN them -> congruent ASA : two angles + the side BETWEEN them -> congruent AAS : two angles + a side not between them -> congruent (via angle-sum) -------------------------------------------------------------- SSA : two sides + a non-included angle -> NOT enough AAA : all three angles, no side -> NOT enough (only similar)
The two that look like rules but aren't
It is tempting to think any three matching parts must lock a triangle, but two arrangements fail, and knowing why protects you from a whole class of wrong proofs. These are the SSA and AAA non-criteria. AAA — all three angles equal — fixes only the shape, not the size: a small triangle and a large one can share all three angles, like a photo and its enlargement. Equal angles give you similar figures, a powerful idea in its own right, but not congruence.
SSA is subtler and trips up more people. You know two sides and an angle not between them. Picture one fixed side, an angle leaving its near end, and a second side of fixed length swinging down from the far ray to meet the baseline. If that swinging side is short enough, it can reach the baseline in two different places — making two genuinely different triangles from the same SSA data. This is the famous ambiguous case: the same three measurements describe two non-congruent triangles, so SSA cannot be a congruence rule.
A worked proof, and the payoff CPCTC
Let us run a real one. Suppose M is the midpoint of both segments AC and BD — the two segments cross at M, each cutting the other in half. Claim: triangle AMB = triangle CMD. We have AM = CM and BM = DM because M bisects each segment. The angles at M, namely angle AMB and angle CMD, are vertical angles, so they are equal — a fact you proved back in the first guide of this rung. That is two sides and the included angle: exactly SAS.
- AM = CM, because M is the midpoint of AC (given).
- angle AMB = angle CMD, because they are vertical angles at the crossing M.
- BM = DM, because M is the midpoint of BD (given).
- By SAS, triangle AMB = triangle CMD.
- Therefore AB = CD, by CPCTC.
That last line is the whole reason proving congruence is worth the trouble. Once two triangles are known congruent, every remaining pair of corresponding parts is automatically equal too — the three pairs you did not use to prove it. This principle is CPCTC: Corresponding Parts of Congruent Triangles are Congruent. We used three matching parts (two sides, one angle) to earn the congruence; CPCTC then hands us the other three for free, which is how we conclude AB = CD without ever measuring AB or CD directly.
How to choose your criterion
In practice, proving a congruence is mostly bookkeeping: collect the equal parts the problem hands you, see how they are arranged, and name the matching criterion. Shared sides count (a side equals itself), midpoints give equal halves, vertical angles give equal angles, and parallel lines from the second guide of this rung hand you equal corresponding angles across a transversal. The skill is reading the arrangement: is the known angle squeezed between the two known sides, or off to the side?
Run through them in order. Have three sides? SSS. Two sides with the angle wedged between? SAS. Two angles with a side between, or two angles and any side? ASA or AAS. A right triangle with hypotenuse and a leg? HL. And if all you have is two sides and a stray angle off in the corner, stop — that is the SSA trap, not a proof. Matching the picture to the right four-letter label is, honestly, ninety percent of triangle proofs at this level.
These four criteria are the toolkit you will reach for again and again. The next guide turns them inward on a single triangle: when two sides of one triangle are equal, an elegant congruence argument proves the base angles must be equal too — the heart of the isosceles triangle and its bisectors. Everything you built here carries straight over.