▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original triangle. Learn more about the controls
Labels toggles the point names, Measures writes each angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Through one vertex draw the line parallel to the opposite side. The two alternate-angle pairs carry the other two angles up to that vertex, where all three fill a straight angle. Drag any vertex: the sum stays 180° for every triangle.Triangle angle sum
Display
Drag A, B or C
α 53.5° + β 50.4° + γ 76.0° = 180.0°
Step 1 / 5
Step explanations
1Triangle ABC
Given Any triangle with interior angles α, β, γ. To prove: α + β + γ = 180°.
Step through the proof, then drag the triangle to see that nothing in the argument depends on its shape. 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.
Proof?Triangle angle sumDisplayLabelsMeasuresDrag A, B or C
α 53.5° + β 50.4° + γ 76.0° = 180.0°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 5
Step explanations
1Triangle ABC
Given Any triangle with interior angles α, β, γ. To prove: α + β + γ = 180°.
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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 angle's size beside its arc
3Solid points — drag A, B or C at any step; the hollow points D and E are constructed and follow
4Readout — the line under the figure: the three angles and their sum, live while you drag; the sum stays at 180°
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original triangle
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
The Statement
For every triangle ABC with interior angles α at A, β at B and γ at C:
α+β+γ=180°
The claim is about all triangles at once. The proof never uses a measurement, which is why dragging the vertices cannot break it.
Opening scene, frozen
Any triangle with its three interior angles marked α, β, γ.
The Parallel Line
Through C draw the line DE parallel to AB. A parallel through C can be constructed without any extra assumption; the parallel postulate adds that it is the only one. The next step, equal alternate angles, depends on that uniqueness. In geometries where the postulate fails, such as on a sphere, the angle sum of a triangle is not 180°.
Parallel line, frozen
Line DE through C, parallel to AB.
Alternate Angles
AC crosses both parallel lines, so the angle it makes with DE at C equals the angle it makes with AB at A: ∠ACD=α. The same holds for BC: ∠BCE=β.
Each original angle is copied to C, on its own side of γ.
Both alternate pairs, frozen
The angles at A and B reappear at C, one on each side of γ.
Why It Holds
At C the angles α, γ and β sit side by side along the straight line DE with no gaps and no overlap. A straight angle measures 180°, so α+γ+β=180°.
Straight angle, frozen
α, γ and β fill the straight angle DCE: 180°.
Consequences
• Each angle of an equilateral triangle is 60° • A triangle has at most one right or obtuse angle • The two acute angles of a right triangle are complementary • An exterior angle equals the sum of the two remote interior angles