▶ 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 copy D follows, and the readout under the canvas keeps the apex angles at C and D equal. 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?Put the second triangle on the first one’s side AB, on the far side, as ABD with AD = AC and BD = BC. Join C to D. Triangle ACD is isosceles, so its base angles at C and D are equal; so is BCD. Adding the two pairs gives ∠ACB = ∠ADB, and then AC = AD, CB = DB with the equal angle between them make the triangles congruent by SAS. Drag any vertex.SSS congruence
Display
Drag A, B or C; the copy D follows
∠ACB = 93.4°, ∠ADB = 93.4°
Step 1 / 6
Step explanations
1Two triangles with three equal sides, base to base
Given ABC and ABD have AB in common, AC = AD (one tick each) and BC = BD (two ticks each). ABD is the second triangle of the theorem, moved so that its side equal to AB lies on AB, on the far side. To prove: ABC ≅ ABD.
Step through the proof, then drag the vertices of the upper triangle and watch its copy below the base keep the same angle at its apex. 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 congruenceDisplayLabelsMeasuresDrag A, B or C; the copy D follows
∠ACB = 93.4°, ∠ADB = 93.4°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Two triangles with three equal sides, base to base
Given ABC and ABD have AB in common, AC = AD (one tick each) and BC = BD (two ticks each). ABD is the second triangle of the theorem, moved so that its side equal to AB lies on AB, on the far side. To prove: ABC ≅ ABD.
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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 — A, B and C move freely; the hollow apex D of the copy is placed with AD = AC and BD = BC
4Readout — the line under the figure: the apex angles at C and 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?SSS congruenceDisplayLabelsMeasuresDrag A, B or C; the copy D follows
∠ACB = 93.4°, ∠ADB = 93.4°
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Two triangles with three equal sides, base to base
Given ABC and ABD have AB in common, AC = AD (one tick each) and BC = BD (two ticks each). ABD is the second triangle of the theorem, moved so that its side equal to AB lies on AB, on the far side. To prove: ABC ≅ ABD.
2Join C to D
Construction The segment CD crosses AB and splits each of the two triangles. It also makes two new triangles that straddle AB: ACD, with AC = AD, and BCD, with BC = BD. Both are isosceles.
3∠ACD = ∠ADC = α
Isosceles base angles In triangle ACD the sides AC and AD are equal, so the base angles at C and at D are equal.
4∠BCD = ∠BDC = β
Isosceles base angles In triangle BCD the sides BC and BD are equal, so the base angles at C and at D are equal too.
5∠ACB = ∠ADB
Adding the parts CD runs inside both angles, so ∠ACB = α + β and ∠ADB = α + β: the angles at C and at D are equal.
6ABC ≅ ABD
SAS congruence AC = AD, CB = DB, and the angles between those sides, at C and at D, have just been shown equal. Two sides and the included angle: the triangles are congruent by SAS. Since ABD is a copy of the second triangle, the two given triangles are congruent. ∎
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1The two triangles — ABC above and ABD below the common side, both lit at the last step
2The angles at C — α and β, the two pieces made by CD; their sum is the apex angle
3The angles at D — the same α and β, from the two isosceles triangles
4Justification tags — Given, Construction, Isosceles base angles (twice), Adding the parts, 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: all three sides and all three angles agree. The congruence criteria say that less is enough. Side-angle-side, used on many of these pages, says two sides and the angle between them determine a triangle; this page proves the side-side-side criterion from it:
AB=A′B′,BC=B′C′,CA=C′A′⟹ABC≅A′B′C′
three pairs of equal sides force the angles to agree too. It is Proposition 8 of Book I of Euclid's Elements. Euclid proved it by laying one triangle on the other; the proof here, found by Philo of Byzantium, draws the second triangle on the far side of a common side and uses the isosceles base angles theorem twice instead.
In the player the lower triangle is the copy: its apex D is placed so that AD=AC and BD=BC, and it follows when A, B or C is dragged. The readout under the canvas shows the apex angles at C and at D, which the proof shows to be equal.
Placing the Second Triangle
Take the common side first. The second triangle has a side A′B′ equal to AB; move it so that this side lies on AB with A′ at A and B′ at B, and so that its third vertex falls on the other side of the line AB from C. Call that vertex D. Then AD=A′C′=AC and BD=B′C′=BC, marked in the player with one tick and two ticks.
The two triangles now sit base to base on AB, one above and one below, filled in two colours. Nothing has been assumed about their angles. The claim to prove is that the angle at D equals the angle at C; once it does, side-angle-side applies and the triangles are congruent.
Opening scene, frozen
ABC above and its copy ABD below the common side: AC = AD, BC = BD.
Joining C and D
The construction is one segment, CD. It crosses the common side AB and splits each triangle into two. More usefully, it creates two new triangles that straddle AB: the triangle ACD, whose sides AC and AD are equal, and the triangle BCD, whose sides BC and BD are equal. Both are isosceles, with their apexes at A and at B and their common base CD.
Everything the proof needs is now in the picture: the angles at C and D have each been cut into two pieces by CD, and the pieces belong to isosceles triangles.
Joining C and D, frozen
The segment CD across AB: two isosceles triangles ACD and BCD.
The First Isosceles Triangle
In triangle ACD, AC=AD, so by the isosceles base angles theorem the angles at the ends of the base CD are equal:
∠ACD=∠ADC=α
The player marks both with the same colour. ∠ACD is the piece of the angle at C on the side of A; ∠ADC is the piece of the angle at D on the side of A.
The first isosceles triangle, frozen
In ACD the base angles at C and D are both α.
The Second Isosceles Triangle
In triangle BCD, BC=BD, so the same theorem gives
∠BCD=∠BDC=β
These are the other two pieces: the piece of the angle at C on the side of B, and the piece of the angle at D on the side of B. Two isosceles triangles, two pairs of equal base angles, and the four pieces are accounted for.
The second isosceles triangle, frozen
In BCD the base angles at C and D are both β.
Adding the Angles
The segment CD passes through the inside of the angle at C and the inside of the angle at D, so each whole angle is the sum of its two pieces:
∠ACB=α+β,∠ADB=α+β
So ∠ACB=∠ADB: the apex angles of the two triangles are equal. This is the step the whole proof was for. (When CD falls outside one of the angles, which happens if the angle at A or B is obtuse, the whole angle is a difference β−α instead of a sum, and the argument is the same; the player keeps to the case drawn.)
Adding the angles, frozen
α + β at C and α + β at D: the apex angles are equal.
Why It Holds
Now compare the two triangles with side-angle-side. AC=AD, CB=DB, and the angles between those sides, at C and at D, have just been shown equal. So ABC≅ABD. And ABD is the second triangle moved into place, so ABC≅A′B′C′: three equal sides make congruent triangles. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given sides, a construction, the isosceles theorem twice, an addition, and side-angle-side.
The proof reduces one criterion to another. Side-angle-side itself is not proved on these pages; Euclid proves it by superposition, laying one triangle on the other, and most modern treatments take it as an axiom. Everything else, including this page, is built on it.
The conclusion, frozen
Two sides and the included angle agree: ABC ≅ ABD by SAS.
Dragging the Figure
Drag C: the upper triangle changes shape and the lower one changes with it, since D is always placed with AD=AC and BD=BC; the readout keeps the two apex angles equal. Drag A or B to change the common side. Watch the pieces α and β trade sizes as C moves from side to side while the sums stay equal.
The player keeps the angles at A and B acute, so that CD crosses AB between A and B and runs inside both apex angles; a drag that would make one of those angles obtuse, or carry a vertex out of the frame, is refused. The theorem holds in the obtuse case too, with a difference of angles in place of a sum.
After dragging C, frozen
C pulled right: α and β have traded sizes in both triangles, and the apex angles are still equal.
Consequences
• Rigidity. A triangle with given sides has only one shape, which is why triangles brace frames, bridges and roofs: three bars pinned together cannot flex, while four can. • Three sides, one triangle. The construction of a triangle from three given lengths, with two compass arcs, gives the same triangle whoever draws it, provided the lengths satisfy the triangle inequality. • Used on these pages. Side-side-side carried the rhombus diagonals page and the third bisector on the circumcentre page. • ASA and AAS. Angle-side-angle also follows from side-angle-side, by contradiction: if the sides next to the equal angles differed, cutting the longer one down to the shorter would give, by SAS, a second triangle with the same angle at the far vertex, which is impossible. Angle-angle-side then follows from the angle sum. • Right triangles. Hypotenuse and one leg determine a right triangle, since the Pythagorean theorem gives the other leg and SSS finishes.