What does the SAS similarity criterion say?

If A'B'/AB = A'C'/AC and the included angles at A and A' are equal, then the triangles ABC and A'B'C' are similar: all three angles are equal and B'C'/BC is the same ratio.

How is SAS similarity proved?

Mark D on AB and E on AC with AD = A'B' and AE = A'C', so AD/AB = AE/AC. Same heights give [ADE]/[ABE] = AD/AB and [ADE]/[ACD] = AE/AC, so [ABE] = [ACD] and [BDE] = [CDE]; equal areas on the same base DE put B and C at the same distance from DE, so DE is parallel to BC. Corresponding angles and AA finish.

Why must the equal angle be the included one?

Two sides in ratio with an equal angle opposite one of them do not determine the triangle: the third vertex can be in two positions, just as with the failed SSA congruence. With the angle between the two sides the triangle is fixed.

SAS 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 ratios 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′ = ∠A and A′B′/AB = A′C′/AC. Cut off AD = A′B′ and AE = A′C′, so AD/AB = AE/AC and ADE ≅ A′B′C′ by SAS. Same heights give [ADE]/[ABE] = AD/AB and [ADE]/[ACD] = AE/AC, so [ABE] = [ACD] and, taking away ADE, [BDE] = [CDE]: equal areas on the same base DE put B and C at the same distance from DE, so DE ∥ BC. Corresponding angles and AA finish. Drag D to resize the second triangle, or a vertex.SAS 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 / 6

Step explanations

1Two sides in ratio, included angle equal

Given
A second triangle A′B′C′ (not drawn) has ∠A′ = ∠A = α and A′B′/AB = A′C′/AC. On AB mark D with AD = A′B′ (one tick) and on AC mark E with AE = A′C′ (two ticks), so that AD/AB = AE/AC. By SAS, ADE ≅ A′B′C′. To prove: ADE, and so A′B′C′, is similar to ABC.

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 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

DE

stay parallel to

BC

and 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?SAS 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 / 6

Step explanations

1Two sides in ratio, included angle equal

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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 AE/AC = AD/AB

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 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?SAS 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 6 / 6

Step explanations

1Two sides in ratio, included angle equal

2AD/AB = [ADE] / [ABE]

Same height
Join B to E. Triangles ADE and ABE have their bases AD and AB on the line AB and the same apex E, so their areas are in the ratio of their bases.

3AE/AC = [ADE] / [ACD]

Same height
Join C to D. Triangles ADE and ACD have their bases AE and AC on the line AC and the same apex D, so [ADE] / [ACD] = AE/AC.

4[BDE] = [CDE]

Equal ratios
AD/AB = AE/AC was given, so [ADE] / [ABE] = [ADE] / [ACD], and therefore [ABE] = [ACD]. Both contain the triangle ADE; take it away from each: [BDE] = [CDE].

5DE ∥ BC

Same base
Triangles BDE and CDE have the same base DE and equal areas, so they have equal heights: B and C are at the same distance from the line DE, and on the same side of it. A line through two such points is parallel to DE: BC ∥ DE.

6ADE ~ ABC

AA similarity
With DE ∥ BC the corresponding angles are equal: ∠ADE = ∠B = β and ∠AED = ∠C = γ. The triangles ADE and ABC have equal angles, so by AA similarity DE/BC = AD/AB = AE/AC. Since ADE ≅ A′B′C′, the same holds for A′B′C′. ∎

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

2The cross lines — BE and CD: the triangles ABE and ACD have equal areas

3The parallel found — DE and BC, marked with the same chevron

4Justification tags — Given, Same height (twice), Equal ratios, Same base, AA similarity

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

The Statement

The SAS similarity criterion says: if two sides of one triangle are in the same ratio to two sides of another, and the angles between those sides are equal, the triangles are similar. With

\angle A' = \angle A

and

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

all three angles are equal and the third pair of sides is in the same ratio.

As on the AA similarity page the second triangle is not drawn. Mark

D

AB

with

AD = A'B'

and

E

AC

with

AE = A'C'

(one and two ticks); by SAS congruence the triangle

ADE

is a congruent copy of

A'B'C'

, and the hypothesis says

AD/AB = AE/AC

. What has to be shown is that

ADE

is similar to

ABC

, and the key is to prove that

DE

is parallel to

BC

: the converse of the intercept theorem, proved here by areas.

Opening scene, frozen

ADE shaded: AD = A′B′, AE = A′C′, so AD/AB = AE/AC and ADE ≅ A′B′C′.

Cutting the Side AB

Join

B

E

. The triangles

ADE

and

ABE

have their bases

AD

and

AB

on the same line and the same apex

E

, so they have the same height:

\frac{[ADE]}{[ABE]} = \frac{AD}{AB}

Cutting the side AB, frozen

ADE and ABE with the apex E: [ADE]/[ABE] = AD/AB.

Cutting the Side AC

Join

C

D

. The triangles

ADE

and

ACD

have their bases

AE

and

AC

on the line

AC

and the same apex

D

\frac{[ADE]}{[ACD]} = \frac{AE}{AC}

Cutting the side AC, frozen

ADE and ACD with the apex D: [ADE]/[ACD] = AE/AC.

Equal Areas

The hypothesis

AD/AB = AE/AC

now reads

[ADE]/[ABE] = [ADE]/[ACD]

. The numerators are the same, so the denominators are equal:

[ABE] = [ACD]

. Each of these two triangles is made of

ADE

and one more piece,

BDE

CDE

. Take

ADE

away from both:

[BDE] = [CDE]

Equal areas, frozen

BDE and CDE shaded: [ABE] = [ACD], so [BDE] = [CDE].

The Parallel

The triangles

BDE

and

CDE

have the same base

DE

and equal areas, so their heights are equal:

B

and

C

are at the same distance from the line

DE

, and on the same side of it. A line through two points at the same distance from a line, on the same side, is parallel to it:

BC \parallel DE

This is the converse of the intercept theorem, and the argument is that theorem's proof read backwards.

The parallel, frozen

Same base, equal areas, equal heights: DE ∥ BC.

Why It Holds

With

DE \parallel BC

the corresponding angles are equal (parallel lines):

\angle ADE = \angle B = \beta

and

\angle AED = \angle C = \gamma

. The triangles

ADE

and

ABC

share the angle at

A

and now agree in all three angles, so by AA similarity

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

Since

ADE

is a congruent copy of

A'B'C'

, the same angles and the same ratio hold for

A'B'C'

. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, same-height triangles twice, equal ratios, equal areas on the same base, and AA similarity.

The conclusion, frozen

Corresponding angles β and γ: ADE ~ ABC by AA.

Dragging the Figure

Drag

D

along

AB

E

moves along

AC

so that

AD/AB = AE/AC

stays true (that is the hypothesis), the copy

ADE

grows or shrinks, and

DE

stays parallel to

BC

, as the proof says it must. The readout shows the common ratio,

DE/BC

, and the angles at

D

and at

B

, which stay equal. Drag

A

B

C

to reshape the big triangle.

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, DE still parallel to BC.

Consequences

• The converse of the intercept theorem. The middle of this proof shows that a line cutting two sides of a triangle in the same ratio is parallel to the third side; it is used on its own as often as the theorem itself.
• SSS similarity. The SSS similarity criterion is proved by SAS similarity in two lines.
• Scaling from a point. Enlarging a figure from a centre

A

by a factor

k

sends every segment to a parallel segment

k

times as long; SAS similarity is the reason, applied to each triangle with a vertex at

A

.
• The midsegment theorem. When

D

and

E

are midpoints, the hypothesis holds with ratio

\tfrac12

and the conclusion is the midsegment theorem, both the parallel and the half length.
• Spiral similarity. With the included angle equal but the triangles turned, SAS similarity is the criterion behind spiral similarities, which turn and scale a figure about one centre.