What does the AA similarity criterion say?

If two angles of one triangle equal two angles of another, the triangles are similar: the third angles are equal and the three pairs of corresponding sides are in the same ratio.

How is AA similarity proved?

Cut off AD = A'B' on AB and draw DE parallel to BC. Corresponding angles give angle ADE = angle B, so ADE is congruent to A'B'C' by ASA. The intercept theorem gives AD/AB = AE/AC, and a second parallel EF through E, with DEFB a parallelogram, gives AE/AC = DE/BC.

Why is one equal angle not enough for similarity?

Fix an angle and change the lengths of its two arms independently: the triangles you get have the same angle at that vertex but different shapes. With two equal angles the third is fixed by the angle sum, and the shape is determined.

AA 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 three ratios and the angle at $D$ . 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 second triangle A′B′C′ (not drawn) has ∠A′ = α and ∠B′ = β. Cut off AD = A′B′ on AB and draw DE ∥ BC. Corresponding angles make ∠ADE = β, so ADE ≅ A′B′C′ by ASA. The intercept theorem gives AD/AB = AE/AC, and with EF ∥ AB (DEFB a parallelogram) also DE/BC; replace AD, AE, DE by the sides of A′B′C′. Drag D to resize the second triangle, or a vertex.AA similarity

DisplayLabelsMeasures

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

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

Step 1 / 6

Step explanations

1Two equal angles

Given
Triangle ABC has the angles α at A and β at B. A second triangle A′B′C′ (not drawn) has the same two angles, ∠A′ = α and ∠B′ = β. On AB mark the point D with AD = A′B′ (the tick). To prove: A′B′/AB = B′C′/BC = C′A′/CA, the triangles are similar.

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

Getting Started with the Visualizer

The Statement

The Parallel Through D

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

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 three ratios in the readout 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?AA similarityDisplayLabelsMeasuresDrag D along AB (the size of the second triangle), or A, B or C

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

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1Two equal angles

1234567

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 points E and F follow

4Readout — the line under the figure: AD/AB, AE/AC, DE/BC and the angle at D, 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 6 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 6 of 6

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

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

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1Two equal angles

2∠ADE = β

Parallel lines
Through D draw the parallel to BC; it meets AC at E. The line AB is a transversal of the parallels DE and BC, so the corresponding angles are equal: ∠ADE = ∠ABC = β.

3ADE ≅ A′B′C′

ASA
Triangles ADE and A′B′C′ have ∠A = ∠A′ = α, the side AD = A′B′ between the angles, and ∠ADE = ∠B′ = β. By ASA they are congruent, so AE = A′C′ and DE = B′C′.

4AD/AB = AE/AC

Intercept theorem
DE ∥ BC, so by the intercept theorem AD/DB = AE/EC. Flip both fractions and add 1: (DB + AD)/AD = (EC + AE)/AE, that is AB/AD = AC/AE, or AD/AB = AE/AC.

5AE/AC = DE/BC

Intercept theorem
Through E draw the parallel to AB; it meets BC at F. In DEFB both pairs of opposite sides are parallel, so it is a parallelogram and DE = BF. The intercept theorem in the angle at C, with EF ∥ AB, gives CE/CA = CF/CB, hence AE/AC = BF/BC, and AE/AC = DE/BC.

6A′B′/AB = A′C′/AC = B′C′/BC

Substituting
So AD/AB = AE/AC = DE/BC. Replace AD, AE and DE by their equals A′B′, A′C′ and B′C′: the three sides of A′B′C′ are in one ratio to the three sides of ABC. The third angles agree too, since both triangles have angle sum 180°. ∎

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1The copy ADE — the shaded triangle, congruent to A′B′C′ by ASA

2The two parallels — DE with one chevron like BC, EF with two like AB

3The parallelogram DEFB — shaded below the copy: DE = BF

4Justification tags — Given, Parallel lines, ASA, Intercept theorem (twice), Substituting

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

The Statement

Two triangles are similar when their angles are equal and their sides are in proportion. The AA criterion says that two equal angles are enough: if triangles

ABC

and

A'B'C'

have

\angle A' = \angle A

and

\angle B' = \angle B

, then

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

The third angles are equal at once, by the angle sum; the content of the theorem is the proportion of the sides.

The second triangle is not drawn in the player. Instead its size is recorded by the point

D

AB

with

AD = A'B'

(the tick), and the proof builds a congruent copy

ADE

A'B'C'

inside

ABC

. Drag

D

to choose any size for

A'B'C'

; the readout shows the three ratios, all equal to

AD/AB

Opening scene, frozen

The angles α and β of ABC; AD = A′B′ marked on AB.

The Parallel Through D

Through

D

draw the line parallel to

BC

; it meets

AC

E

. The side

AB

is a transversal of the two parallels, so the corresponding angles are equal (parallel lines and a transversal):

\angle ADE = \angle ABC = \beta

The small triangle

ADE

now has the angle

\alpha

A

and the angle

\beta

D

, the same two angles as

A'B'C'

The parallel through D, frozen

DE parallel to BC: the corresponding angle at D is β.

The Congruent Copy

Compare

ADE

with

A'B'C'

: the angle at

A

equals the angle at

A'

, the side

AD

equals

A'B'

by construction, and the angle at

D

equals the angle at

B'

. Two angles and the side between them: by ASA the triangles are congruent, and so

AE = A'C', \qquad DE = B'C'

Everything about

A'B'C'

is now inside

ABC

, as the triangle

ADE

The congruent copy, frozen

ADE shaded: α, AD = A′B′, β, so ADE ≅ A′B′C′ by ASA.

The First Two Ratios

DE \parallel BC

, so the intercept theorem gives

AD/DB = AE/EC

. Flip both fractions and add 1 to each side:

(DB + AD)/AD = (EC + AE)/AE

, that is

\frac{AD}{AB} = \frac{AE}{AC}

Two of the three sides of the copy are in the same ratio to the sides of

ABC

The first two ratios, frozen

By the intercept theorem AD/AB = AE/AC.

The Third Ratio

The side

DE

does not lie along a side of

ABC

, so one more construction is needed. Through

E

draw the parallel to

AB

; it meets

BC

F

. In the quadrilateral

DEFB

both pairs of opposite sides are parallel, so it is a parallelogram and its opposite sides are equal:

DE = BF

. Now apply the intercept theorem in the angle at

C

, where

EF \parallel AB

cuts

CA

and

CB

CE/CA = CF/CB

, so

AE/AC = BF/BC

, and

\frac{AE}{AC} = \frac{DE}{BC}

The third ratio, frozen

EF parallel to AB; DEFB is a parallelogram, so DE = BF and AE/AC = DE/BC.

Why It Holds

Put the two results together:

AD/AB = AE/AC = DE/BC

. Replace

AD

AE

DE

by their equals

A'B'

A'C'

B'C'

from the congruent copy:

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

The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, corresponding angles on parallel lines, ASA, the intercept theorem twice, and a substitution.

Once AA is in hand, every other similarity criterion is a short step away: the SAS and SSS similarity pages both end by calling AA.

The conclusion, frozen

AD/AB = AE/AC = DE/BC, with AD, AE, DE the sides of A′B′C′.

Dragging the Figure

Drag

D

along

AB

: the copy

ADE

grows or shrinks, the parallels

DE

and

EF

move with it, and the three ratios in the readout change together and stay equal. When

D

is the midpoint of

AB

the copy is the midsegment triangle and every ratio is

0.50

. Drag

A

B

C

to reshape the big triangle;

D

keeps its fraction of

AB

and the copy keeps its angles.

The player keeps

D

between

A

and

B

, that is, the second triangle smaller than the first. If

A'B'C'

is the larger one, swap the roles of the two triangles; the ratios are then the reciprocals. 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 three ratios still equal.

Consequences

• The other criteria. SAS similarity and SSS similarity reduce to AA once a parallel is found.
• Right triangles. Two right triangles with one equal acute angle are similar; the altitude to the hypotenuse cuts a right triangle into two triangles similar to it, which gives the geometric-mean relations and a second proof of the Pythagorean theorem.
• The angle bisector theorem. The angle bisector theorem is often proved by AA, with a parallel to the bisector through a vertex; this section proves it by areas instead.
• Circles. Equal inscribed angles on the same arc make similar triangles; that is the usual route to the intersecting chords and secant theorems, which this section proved by Pythagoras instead, and the route to Ptolemy's theorem.
• Trigonometry. The sine, cosine and tangent of an angle are well defined because all right triangles with that angle are similar by AA.