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The ASA Congruence Criterion

How to use
  1. ▶ 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
  2. Slide EE along ACAC, or drag AA, BB or CC; the readout under the canvas compares the angle ABEABE with the angle ABCABC. Learn more about why it holds
  3. 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?Suppose the sides AC and A′C′ next to the equal angles were different, say AC longer. Cut off AE = A′C′ on AC. Then ABE and A′B′C′ agree in two sides and the included angle, so by SAS the angle ABE equals the angle at B′, which equals the angle ABC. But E lies inside the angle ABC, so ∠ABE is only a part of it. The sides cannot differ, and SAS finishes. Slide E along AC to see the part fall short of the whole.ASA congruence
Display
Slide E along AC; drag A, B or C to reshape the triangle
αβABC
∠ABC = 50.7°, ∠ABE = 19.9°
Step 1 / 6

Step explanations

1Two angles and the side between them
Given
Triangle ABC has angle α at A, angle β at B, and the side AB between them (one tick). A second triangle A′B′C′ has the same angle α at A′, the same angle β at B′, and A′B′ = AB. To prove: ABC ≅ A′B′C′.
The second triangle is not drawn; E carries its side A′C′. Learn more about the statement

Getting Started with the Visualizer
The Statement
Suppose the Sides Differ
Two Sides and the Angle Between
A Part of the Angle
The Contradiction
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then slide the point EE along the side ACAC and watch the angle ABEABE fall short of the angle ABCABC for every position. 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
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 — A, B and C move freely; from step 2 the point E slides along AC
4Readout — the line under the figure: the angle ABC and, from step 2, the angle ABE, 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
12345
1The point E — on AC with AE = A′C′ (two ticks); slide it along the side
2The triangle ABE — shaded: congruent to the second triangle by SAS under the supposition
3The angles at B — β, the given angle, and β′, the angle ABE: a part of it
4Justification tags — Given, Supposition, SAS congruence, A part of the whole, Contradiction, SAS congruence
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Two triangles are congruent when one can be laid exactly on the other. The congruence criteria say which three measurements are enough to force that. Side-angle-side is the one taken as given on these pages; the SSS congruence page derived side-side-side from it. This page derives angle-side-angle:

∠A=∠A′,  AB=A′B′,  ∠B=∠B′  ⟹  ABC≅A′B′C′\angle A = \angle A', \; AB = A'B', \; \angle B = \angle B' \implies ABC \cong A'B'C'


two angles and the side between them determine a triangle. It is the first half of Proposition 26 of Book I of Euclid's Elements, and the proof is Euclid's: suppose the triangles were not congruent, and find a part equal to its whole.

The player shows the first triangle ABCABC with its angle α\alpha at AA, its angle β\beta at BB and the side ABAB between them marked with one tick. The second triangle is not drawn; everything known about it is carried by the point EE, which is placed on ACAC so that AEAE equals the second triangle's side A′C′A'C'. The readout under the canvas shows the angles ABCABC and ABEABE.
αβABC
Opening scene, frozen

Triangle ABC with the angle α at A, the angle β at B and the side AB between them.


Suppose the Sides Differ

The whole proof comes down to one length. If AC=A′C′AC = A'C' then the two triangles agree in AB=A′B′AB = A'B', AC=A′C′AC = A'C' and the angle α\alpha between those sides, and side-angle-side makes them congruent at once. So it is enough to show AC=A′C′AC = A'C'.

The proof is by contradiction, the same shape of argument as on the tangent perpendicular to radius page. Suppose the two sides are not equal. Then one of them is longer; the two triangles play the same role, so there is no loss in supposing ACAC is the longer. The shorter length A′C′A'C' then fits inside ACAC: measure it off from AA along ACAC and call the end EE. The player marks AEAE with two ticks, draws EE as a draggable point strictly between AA and CC, and joins BB to EE.
αβABCE
Supposing the sides differ, frozen

E on AC with AE = A′C′ (two ticks), and the segment BE.


Two Sides and the Angle Between

Compare the new triangle ABEABE with the second triangle A′B′C′A'B'C':

AB=A′B′,AE=A′C′,∠BAE=∠B′A′C′=αAB = A'B', \qquad AE = A'C', \qquad \angle BAE = \angle B'A'C' = \alpha


The first is given, the second is how EE was chosen, and the third holds because EE lies on ACAC, so the angle BAEBAE is the angle BACBAC itself. Two sides and the included angle agree, so by side-angle-side ABE≅A′B′C′ABE \cong A'B'C'. Congruent triangles have equal corresponding angles, and the angle at BB in ABEABE corresponds to the angle at B′B':

∠ABE=∠A′B′C′=β\angle ABE = \angle A'B'C' = \beta


The player shades the triangle ABEABE and marks its angle at BB as β′\beta': the angle that the supposition says equals β\beta.
αββ′ABCE
Two sides and the angle between, frozen

ABE shaded: congruent to A′B′C′ by SAS, so its angle at B, marked β′, would equal β.


A Part of the Angle

Now look at the picture instead of the supposition. EE lies on the segment ACAC strictly between AA and CC, so the ray from BB through EE runs inside the angle ABCABC, between the rays BABA and BCBC. The angle ABEABE is therefore a part of the angle ABCABC, and a part is less than the whole:

∠ABE<∠ABC=β\angle ABE < \angle ABC = \beta


The readout makes the same comparison in degrees for the figure on screen, and the inequality holds wherever EE is slid along ACAC.
αββ′ABCE
A part of the angle, frozen

BE runs inside the angle ABC: β′ is only a part of β.


The Contradiction

The two previous sections say opposite things about the same angle. From the supposition, ∠ABE=β\angle ABE = \beta. From the picture, ∠ABE<β\angle ABE < \beta. An angle cannot both equal β\beta and be smaller than β\beta, so the supposition was false: ACAC is not longer than A′C′A'C'.

The supposition also covered the other case. If A′C′A'C' were the longer side, the same construction in the second triangle, cutting A′E′=ACA'E' = AC off A′C′A'C', gives A′B′E′≅ABCA'B'E' \cong ABC by side-angle-side and the same impossible equality of a part with its whole. So neither side is longer than the other, and AC=A′C′AC = A'C'.
αββ′ABCE
The contradiction, frozen

β′ cannot both equal β and be smaller: the supposition fails, AC = A′C′.


Why It Holds

With AC=A′C′AC = A'C' established, the two triangles agree in AB=A′B′AB = A'B', AC=A′C′AC = A'C' and the angle α\alpha between those sides, and side-angle-side gives ABC≅A′B′C′ABC \cong A'B'C'. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given data, a supposition, side-angle-side, a part being less than its whole, the contradiction, and side-angle-side again.

The shape of the argument is worth noticing. Nothing is measured and nothing is constructed in the given triangles; the only construction, the point EE, lives inside the supposition, and the proof shows that it cannot exist. Side-angle-side is used twice, once to turn the supposition into an equality of angles and once, after the supposition has collapsed, to finish.
αββ′ABCE
The conclusion, frozen

AB, AC and the angle α between them agree: ABC ≅ A′B′C′ by SAS.


Dragging the Figure

Slide EE along ACAC: the shaded triangle ABEABE grows and shrinks, and the readout shows the angle ABEABE staying below the angle ABCABC the whole way, approaching it only as EE approaches CC. Drag AA, BB or CC to reshape the triangle; EE keeps its place as a fraction of the way along ACAC.

The player keeps EE strictly between AA and CC, away from both ends, so that the part is always a visible part; a drag that would carry a vertex out of the frame or flatten the triangle is refused.
αββ′ABCE
After dragging E and C, frozen

E slid towards C and C pulled right: a larger part of a different whole, still short of it.


Consequences

• Used on these pages. Angle-side-angle carried the parallelogram opposite sides page, where a diagonal and two pairs of alternate angles make two congruent triangles, and the parallelogram diagonals page.
• AAS. Two angles and a side not between them also determine a triangle: by the angle sum the third angle is determined too, and then the given side is between two known angles. This is the second half of Euclid's Proposition 26.
• The converse of the isosceles theorem. If two angles of a triangle are equal, the sides opposite them are equal: the triangle is congruent to itself read backwards, by angle-side-angle on the base.
• Surveying. A length that cannot be reached, across a river or up a cliff, is fixed by one measured base and two sighted angles, because angle-side-angle leaves the triangle no freedom.
• Three criteria. Side-angle-side, side-side-side and angle-side-angle are the three criteria of Euclid's first book; together with angle-angle-side they cover every way three independent measurements can fix a triangle, apart from the right-triangle case of hypotenuse and leg.