What does the SSS similarity criterion say?

If A'B'/AB = B'C'/BC = C'A'/CA, the triangles ABC and A'B'C' are similar: their corresponding angles are equal.

Is SSS similarity the same as SSS congruence?

SSS congruence needs equal sides and gives equal triangles; SSS similarity needs sides in one common ratio and gives equal angles. Congruence is the case where the ratio is 1.

SSS Similarity

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag $D$ along $AB$ to resize the second triangle, or $A$ , $B$ or $C$ ; the readout under the canvas shows the common ratio and the angles at $D$ and $B$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?A′B′C′ (not drawn) has A′B′/AB = B′C′/BC = C′A′/CA. Cut off AD = A′B′ on AB and AE = A′C′ on AC; then AD/AB = AE/AC with the common angle at A, so ADE ~ ABC by SAS similarity and DE/BC = AD/AB = B′C′/BC, giving DE = B′C′. Now ADE ≅ A′B′C′ by SSS, and its angles are the angles of ABC. Drag D to resize the second triangle, or a vertex.SSS similarity

DisplayLabelsMeasures

Drag D along AB (the size of the second triangle), or A, B or C

AD/AB = AE/AC = 0.45,  DE/BC = 0.45,  ∠ADE = 56°,  ∠B = 56°

Step 1 / 5

Step explanations

1Three sides in the same ratio

Given
A second triangle A′B′C′ (not drawn) has A′B′/AB = B′C′/BC = C′A′/CA. On AB mark D with AD = A′B′ (one tick) and on AC mark E with AE = A′C′ (two ticks); then AD/AB = AE/AC, the given ratio. To prove: the angles of A′B′C′ equal those of ABC, the triangles are similar.

Drag D: the size of the second triangle. Learn more about the statement

Getting Started with the Visualizer

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 6

Step through the proof, then drag

D

along the side

AB

to change the size of the second triangle, or drag a vertex, and watch the angles at

D

and

B

stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

Proof?SSS similarityDisplayLabelsMeasuresDrag D along AB (the size of the second triangle), or A, B or C

AD/AB = AE/AC = 0.45,  DE/BC = 0.45,  ∠ADE = 56°,  ∠B = 56°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 5

Step explanations

1Three sides in the same ratio

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1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — D slides along AB and sets the size of the second triangle; A, B and C move freely; the hollow point E follows with the same ratio

4Readout — the line under the figure: the common ratio, DE/BC and the angles at D and B, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 5 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 5 of 5

Proof?SSS similarityDisplayLabelsMeasuresDrag D along AB (the size of the second triangle), or A, B or C

AD/AB = AE/AC = 0.45,  DE/BC = 0.45,  ∠ADE = 56°,  ∠B = 56°

← Back↻ ReplayNext →ResetNormal ▾Step 5 / 5

Step explanations

1Three sides in the same ratio

2ADE ~ ABC

SAS similarity
Triangles ADE and ABC share the angle at A, and the sides about it are in the same ratio, AD/AB = AE/AC. By SAS similarity, ADE ~ ABC: DE ∥ BC and DE/BC = AD/AB.

3DE = B′C′

Substituting
DE/BC = AD/AB = A′B′/AB, and by hypothesis A′B′/AB = B′C′/BC. So DE/BC = B′C′/BC, that is DE = B′C′ (three ticks).

4ADE ≅ A′B′C′

SSS
The triangles ADE and A′B′C′ now have three pairs of equal sides: AD = A′B′, AE = A′C′, DE = B′C′. By SSS they are congruent, so the angles of A′B′C′ are the angles of ADE.

5∠A′ = α, ∠B′ = β, ∠C′ = γ

Corresponding angles
DE ∥ BC, so the corresponding angles are equal: ∠ADE = ∠B = β and ∠AED = ∠C = γ, and the angle at A is shared. The angles of ADE, and so of A′B′C′, are the angles of ABC: A′B′C′ ~ ABC, with the given ratio of sides. ∎

12345

1The copy ADE — the shaded triangle at A, with one, two and three ticks on its sides

2The parallel — DE and BC, marked with the same chevron

3The angles — β at B and at D, γ at C and at E, α at A

4Justification tags — Given, SAS similarity, Substituting, SSS, Corresponding angles

5Replay — at the last step Play becomes Replay and Next is off

The Statement

The SSS similarity criterion says: if the three sides of one triangle are in the same ratio to the three sides of another,

\frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{C'A'}{CA}

then the triangles are similar, that is, their angles are equal too. It is the similarity version of SSS congruence, and the proof ends by calling that theorem.

As on the AA and SAS pages the second triangle is not drawn: mark

D

AB

with

AD = A'B'

and

E

AC

with

AE = A'C'

. Then

AD/AB = AE/AC

is the given ratio, and the triangle

ADE

has two sides of

A'B'C'

about the angle

A

. The proof shows that its third side is the third side of

A'B'C'

as well.

Opening scene, frozen

ADE shaded: AD = A′B′ (one tick), AE = A′C′ (two ticks).

SAS Similarity First

The triangles

ADE

and

ABC

share the angle at

A

, and the sides about it are in the same ratio,

AD/AB = AE/AC

. By SAS similarity,

ADE

is similar to

ABC

DE

is parallel to

BC

and

\frac{DE}{BC} = \frac{AD}{AB}

SAS similarity first, frozen

Common angle at A, AD/AB = AE/AC: ADE ~ ABC, DE ∥ BC.

The Third Side

Now compare with the hypothesis.

AD/AB = A'B'/AB

by construction, and the hypothesis says

A'B'/AB = B'C'/BC

. So

\frac{DE}{BC} = \frac{B'C'}{BC}, \qquad DE = B'C'

The copy

ADE

has all three sides of

A'B'C'

AD = A'B'

AE = A'C'

and now

DE = B'C'

(three ticks).

The third side, frozen

DE/BC = AD/AB = B′C′/BC, so DE = B′C′ (three ticks).

The Congruent Copy

Three pairs of equal sides: by SSS congruence the triangles

ADE

and

A'B'C'

are congruent, so the angles of

A'B'C'

are the angles of

ADE

\angle A' = \angle DAE, \qquad \angle B' = \angle ADE, \qquad \angle C' = \angle AED

The congruent copy, frozen

Three equal sides: ADE ≅ A′B′C′.

Why It Holds

DE \parallel BC

, so the corresponding angles are equal (parallel lines):

\angle ADE = \angle B = \beta

and

\angle AED = \angle C = \gamma

, while the angle at

A

is shared. The angles of

ADE

, and therefore of

A'B'C'

, are the angles of

ABC

\angle A' = \alpha, \qquad \angle B' = \beta, \qquad \angle C' = \gamma

Together with the given ratio of sides this is the similarity. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, SAS similarity, a substitution, SSS congruence and corresponding angles.

The three criteria now form a chain: AA rests on the intercept theorem, SAS on AA, SSS on SAS, and all of them, in the end, on areas.

The conclusion, frozen

Corresponding angles β and γ: the angles of A′B′C′ are those of ABC.

Dragging the Figure

Drag

D

along

AB

E

moves along

AC

with the same ratio, the copy

ADE

grows or shrinks and keeps its shape,

DE

stays parallel to

BC

, and the angles at

D

and

B

in the readout stay equal. Drag

A

B

C

to reshape the big triangle; the copy follows.

The player keeps the second triangle smaller than the first; for a larger one swap the roles. A drag that would carry a vertex out of the frame or flatten the triangle is refused.

After dragging D, frozen

D further down AB: a larger copy, the same angles.

Consequences

• Shape from sides alone. Three side lengths fix a triangle up to congruence (SSS); three ratios fix it up to similarity. A triangle with sides

3, 4, 5

is right-angled whatever its size.
• Scale drawings and maps. A plan with every length in the same ratio to the original has the same angles everywhere; SSS similarity, applied triangle by triangle, is why a scale drawing is a faithful picture.
• The chain of criteria. AA rests on the intercept theorem, SAS on AA, and SSS on SAS; each proof is a short reduction to the previous one.
• Congruence as a special case. With ratio

1

the three similarity criteria become SSS, SAS and ASA congruence; the proofs here use the congruence criteria and the intercept theorem, nothing more.
• The HL case. For right triangles, hypotenuse and one leg in the same ratio give similarity, by Pythagoras and SSS similarity, parallel to the HL congruence page.