Three matching parts force the other three — as long as they are the right three.
In triangles ABC and DEF
AB = DE = 7 · BC = EF = 9 · ∠B = ∠E = 40°
Which rule proves them congruent?
Two pairs of equal sides with the angle between them — ∠B sits between AB and BC — is exactly SAS. B is not a valid rule at all: with the angle outside the pair, two genuinely different triangles can fit the same measurements.
Six pairs of parts have to match — three sides and three angles.
same shape, same size
every side and angle matches
same shape, any size
angles match, sides only in ratio
You never have to check all six. Three of them, chosen properly, leave the triangle no room to be anything else.
△ABC ≅ △DEF is read straight across: first with first, second with second.
| In △ABC | matches | In △DEF |
|---|---|---|
| ∠A | ↔ | ∠D |
| ∠B | ↔ | ∠E |
| ∠C | ↔ | ∠F |
| AB | ↔ | DE |
| BC | ↔ | EF |
△ABC ≅ △DEF and △ABC ≅ △EDF say different things. If the statement is scrambled, every part you read off it afterwards is wrong — and the exam choices will include exactly that answer.
| Rule | What you need | The catch |
|---|---|---|
| SSS | all three sides | none — three sides lock a triangle |
| SAS | two sides and the angle between them | the angle must be the included one |
| ASA | two angles and the side between them | side sits between the two angles |
| AAS | two angles and a side off to the side | fine, because the third angle is forced |
| HL | hypotenuse and one leg | right triangles only |
Two angles hand you the third for free — they must total 180°. So the moment you have two angles and any one side, the triangle is pinned down.
AB = DE
side
∠B = ∠E
angle — between AB and BC
BC = EF
side
△ABC ≅ △DEF by SAS
side, angle, side in order
B is the letter shared by AB and BC, which is what “included angle” means. Had the given angle been ∠A, this would not be SAS at all.
AAA and SSA. Both look convincing and both fail.
Same angle at A, same side AB, same length BC — and two completely different triangles. That is why SSA proves nothing.
gives similar, not congruent
a photo and its enlargement
the side swings two ways
the ambiguous case
HL is SSA with a right angle — and there the swing is impossible, because a right triangle can only close one way. Use it only when the right angles are actually marked.
CPCTC — corresponding parts of congruent triangles are congruent.
△ABC ≅ △DEF
proved by SAS
∠A ↔ ∠D
first letter with first letter
∠D = 89°
it copies ∠A
CPCTC only comes after the congruence, never before. It is the payoff, not the proof — the three parts you were given do the work, and the other three are collected afterwards.
△KLM ≅ △ZXY, with KL = 11, LM = 8 and ∠M = 64°. Find XY and ∠Y.
K ↔ Z, L ↔ X, M ↔ Y
read straight across
XY ↔ LM, so XY = 8
2nd–3rd letters with 2nd–3rd
∠Y ↔ ∠M, so ∠Y = 64°
3rd with 3rd
Different: the second triangle’s letters are out of alphabetical order. Match by position in the statement, never by how the letters look.
Given ∠A = ∠D, ∠B = ∠E and BC = EF. Which rule?
the side between ∠A and ∠B is AB
shares both letters
BC touches B only
so it is not between them
△ABC ≅ △DEF by AAS
still proves it
Different: the side is off to one side. The triangles are still congruent; only the name changes from ASA to AAS.
△ABC ≅ △RST, with ∠B = (3y + 7)° and ∠S = (5y − 15)°. Find ∠S.
∠B ↔ ∠S, so 3y + 7 = 5y − 15
corresponding parts are equal
22 = 2y, y = 11
− 3y, + 15, ÷ 2
∠S = 5(11) − 15 = 40°
and ∠B = 33 + 7 = 40° ✓
Different: CPCTC turns into an equation. Find y, then substitute — the question wants the angle.
Exam figures state one or two facts and expect you to read the rest off the drawing.
| You see | You get, free |
|---|---|
| two triangles sharing a side | that side equals itself (a free S) |
| two segments crossing | vertical angles are equal (a free A) |
| “M is the midpoint” | the two halves are equal (a free S) |
| “bisects the angle” | the two half-angles are equal (a free A) |
| parallel sides | alternate interior angles are equal (a free A) |
Example: PT and QS cross at R, and R is the midpoint of both. Prove △PRQ ≅ △TRS.
PR = TR
R is the midpoint of PT
∠PRQ = ∠TRS
vertical angles — free
QR = SR
R is the midpoint of QS
△PRQ ≅ △TRS by SAS
the angle at R sits between the sides
Then CPCTC: P ↔ T and Q ↔ S, so PQ = TS. If PQ = 8, TS is 8 too.
A favourite: an isosceles triangle with the segment from the apex to the midpoint of the base. The two halves share that segment, have equal legs and equal half-bases, so they are congruent by SSS. The two angles at the foot are then equal and form a straight line, so each is 90°, and the segment also bisects the apex angle.
UPCAT-type items ask which extra condition completes a given rule, or ask for a side or angle after the congruence. First list the free parts from the figure, then see which single fact finishes SAS, ASA or AAS — and match letters in the stated order before reading off any part. Ten-second check: an “extra condition” that makes SSA is never the answer.
The problem
In △PQR and △STU: ∠P = ∠S, ∠Q = ∠T and PQ = ST. Name the rule.
The equal side PQ lies between the two equal angles, which makes it ASA. Had the side sat outside the pair, it would be AAS. C proves only similarity — matching angles fix the shape but say nothing about the size.
