▶ 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 A, B or C; the readout under the canvas compares the angle at C with 180° minus the angles at A 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?The angle sum of a triangle is 180°, so two angles fix the third: in both triangles the angle at C is 180° minus the two given angles, hence the angles at C and C′ are equal. Now the given side BC lies between two angles known to be equal, β at B and γ at C, and the angle-side-angle criterion makes the triangles congruent. Drag any vertex.AAS congruence
Display
Drag A, B or C to reshape the triangle
∠C = 69.9°, 180° − ∠A − ∠B = 69.9°
Step 1 / 5
Step explanations
1Two angles and a side not between them
Given Triangle ABC has angle α at A, angle β at B, and the side BC (one tick), which touches the angle at B but not the angle at A. A second triangle A′B′C′ has the same angle α at A′, the same angle β at B′, and B′C′ = BC. To prove: ABC ≅ A′B′C′.
Step through the proof, then drag the vertices and watch the readout keep the third angle equal to 180° minus the other two. 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?AAS congruenceDisplayLabelsMeasuresDrag A, B or C to reshape the triangle
∠C = 69.9°, 180° − ∠A − ∠B = 69.9°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 5
Step explanations
1Two angles and a side not between them
Given Triangle ABC has angle α at A, angle β at B, and the side BC (one tick), which touches the angle at B but not the angle at A. A second triangle A′B′C′ has the same angle α at A′, the same angle β at B′, and B′C′ = BC. To prove: ABC ≅ A′B′C′.
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 all move freely
4Readout — the line under the figure: the angle at C beside 180° minus the angles at A 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?AAS congruenceDisplayLabelsMeasuresDrag A, B or C to reshape the triangle
∠C = 69.9°, 180° − ∠A − ∠B = 69.9°
← Back↻ ReplayNext →ResetNormal ▾Step 5 / 5
Step explanations
1Two angles and a side not between them
Given Triangle ABC has angle α at A, angle β at B, and the side BC (one tick), which touches the angle at B but not the angle at A. A second triangle A′B′C′ has the same angle α at A′, the same angle β at B′, and B′C′ = BC. To prove: ABC ≅ A′B′C′.
2α + β + γ = 180°
Triangle angle sum Call the third angle, at C, γ. The three angles of any triangle add up to 180°, so γ = 180° − α − β. The same holds in the second triangle: its angle at C′ is 180° − α − β as well.
3∠C = ∠C′ = γ
Equal remainders Both third angles are the same number, 180° − α − β, so they are equal. The two triangles now agree in all three angles, and in one side: BC = B′C′.
4BC lies between β and γ
ASA congruence The given side BC has the angle β at one end and the angle γ at the other, and B′C′ has the same two angles at its ends. Two angles and the side between them agree: the angle-side-angle criterion applies.
5ABC ≅ A′B′C′
ASA congruence By angle-side-angle the triangles are congruent: two angles and any one side determine a triangle, whether or not the side lies between the angles. ∎
12345
1The given side — BC, one tick: it touches the angle at B but not the angle at A
2The given angles — α at A and β at B, the two angles the second triangle shares
3The third angle — γ at C, found from the angle sum; it puts BC between two known angles
5Replay — at the last step Play becomes Replay and Next is off
The Statement
Two angles and a side determine a triangle whether or not the side lies between the angles. When it does, that is the ASA congruence criterion, proved from side-angle-side on its own page. When it does not, the criterion is called angle-angle-side:
∠A=∠A′,∠B=∠B′,BC=B′C′⟹ABC≅A′B′C′
Here the side BC touches the angle at B but not the angle at A. It is the second half of Proposition 26 of Book I of Euclid's Elements, and the proof is the shortest on these pages: the angle sum turns it into angle-side-angle.
The player shows the triangle ABC with its angle α at A, its angle β at B and the side BC marked with one tick; the second triangle is not drawn, since the argument never leaves the first. The readout under the canvas shows the angle at C and the value 180° minus the two given angles.
Opening scene, frozen
Triangle ABC with the angle α at A, the angle β at B and the side BC, which does not lie between them.
The Angle Sum
Call the third angle of the first triangle, at C, γ. By the triangle angle sum theorem the three angles of any triangle add up to a straight angle:
α+β+γ=180°
so γ=180°−α−β. The second triangle has the same two angles α and β, so its third angle, at C′, is also 180°−α−β. The player marks γ at C and lights the three angles together.
The angle sum, frozen
The third angle γ at C: α + β + γ = 180°, in both triangles.
The Third Angle
Both third angles are the same number, so they are equal:
∠C=∠C′=180°−α−β=γ
The triangles now agree in all three angles. That alone is not enough for congruence, since two triangles of the same shape can differ in size; the one given side is what pins the size down, and it is used next.
The third angle, frozen
γ = 180° − α − β in both triangles, so the angles at C and C′ are equal.
The Side Between Two Angles
Look at the given side again. BC has the angle β at its end B and the angle γ at its end C; B′C′ has the same angle β at B′ and, by the previous step, the same angle γ at C′. So the given side does lie between two equal angles after all, just not the two that were given:
∠B=∠B′,BC=B′C′,∠C=∠C′
This is exactly the angle-side-angle pattern, and the ASA congruence criterion gives ABC≅A′B′C′.
The side between two angles, frozen
BC has β at one end and γ at the other: the ASA pattern.
Why It Holds
The proof has two moves. The angle sum converts the two given angles into the third, so that every angle is known in both triangles; then the given side, whichever it is, lies between two known angles, and angle-side-angle applies. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given data, the angle sum, equal remainders, and angle-side-angle.
The same two moves work when the given side is CA instead of BC, or when the two given angles are at B and C and the side is AB or CA: with all three angles known, any side lies between two of them. That is why angle-angle-side needs no case distinction. What it does need is that the side be the one opposite the same angle in both triangles; a side opposite α in one triangle and opposite β in the other proves nothing.
The conclusion, frozen
ABC ≅ A′B′C′ by ASA.
Dragging the Figure
Drag A, B or C: the triangle changes shape, the three marked angles change with it, and the readout shows the angle at C staying equal to 180° minus the angles at A and B for every shape. Turn on Measures to see all three angles in degrees and add them up.
A drag that would flatten the triangle or carry a vertex out of the frame is refused.
After dragging C, frozen
C pulled left and down: a wider angle at C, and the readout still shows it equal to 180° minus the other two.
Consequences
• Used on these pages. Angle-angle-side carried the angle bisector theorem page, where two right triangles share the bisected angle, and the incentre page, where it makes the distances from the incentre to the sides equal. • Four criteria. With side-angle-side taken as given, these pages have now derived side-side-side, angle-side-angle and angle-angle-side from it. Any two angles and any side, or any two sides and the included angle, or all three sides, determine a triangle. • Why not three angles. Three equal angles give only the same shape, not the same size; that is similarity, not congruence. Angle-angle-side needs its one side for exactly this reason. • Why not SSA. Two sides and an angle not between them do not determine a triangle in general: with AB, BC and the angle at A given, the vertex C can fall at either of two points on a ray from A, giving two different triangles. The exception is a right or obtuse given angle, and the special case of hypotenuse and leg in a right triangle. • Right triangles. A right triangle with one acute angle and any one side is determined, since the right angle is the second angle and angle-angle-side or angle-side-angle applies.