The angle sum and the exterior angle
Tear the three corners off any paper triangle and lay them point to point. They fill a straight line, 180°, whatever triangle you started with.
Tear off the corners and line them up
Draw a triangle on scrap paper, any shape at all, and cut it out. Mark each corner with a dot so you will know which tip is which. Now tear the three corners off and place them side by side so that the three dots meet at one point and the edges touch. The torn pieces close up into a straight line along the bottom. Try it again with a long thin triangle and with a nearly equilateral one. The fit is the same each time. Three corners of a triangle together make a straight angle, 180°.
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Tearing and dragging both point to the answer, but neither proves it. Paper tears unevenly, and a screen rounds each angle to a tenth of a degree, which the panel admits. A test on ten triangles, or a thousand, is still a test on some triangles. A proof has to work for every triangle at once, including ones nobody has drawn.
One parallel line turns the experiment into a proof
The proof in OpenStax, which goes back to Euclid, draws one extra line. Through the top corner C, draw the line parallel to the base AB. Now look at the three angles that sit along that new line at C. The left one and angle A are alternate angles between two parallel lines, so they are equal. The right one and angle B are alternate angles too, so they are equal. The middle one is angle C itself. Those three angles fill a straight line through C, so they add to 180°. Replace the outer two by A and B and you get A + B + C = 180°. Nothing in the argument depends on the shape of the triangle.
The exterior angle is the sum of the two far corners
Extend one side of a triangle past a corner. The angle between the extension and the next side is an exterior angle. At corner C, the exterior angle and the interior angle C sit side by side on a straight line, so exterior angle = 180° − C. But A + B is also 180° − C. Therefore the exterior angle at C equals A + B, the sum of the two interior angles that are not next to it. Euclid states this in the same Proposition 32, before the angle sum. If A = 40° and B = 65°, the exterior angle at C is 105°, and you never needed to find C, which is 75°.
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Take one exterior angle at each corner and add them. Each exterior angle is 180° minus its interior angle, so the three together are 3 × 180° − 180° = 360°. A walker who goes once round a triangular park turns through one full turn. That is Proclus’s corollary for the triangle.
Finding missing angles with algebra
The angle sum turns many picture puzzles into one-line equations. OpenStax gives a triangle whose angles are x, x + 17 and 3x − 62 degrees. Adding them, 5x − 45 = 180, so 5x = 225 and x = 45. The three angles are 45°, 62° and 73°, and they do add to 180°. Always finish with that last check; a slip in the algebra shows up at once. For an isosceles triangle, remember that the two base angles are equal, so an apex angle of 40° leaves (180 − 40) ÷ 2 = 70° for each base angle.
Interior angles: A + B + C = 180°. Exterior angle at C = A + B. One exterior angle at each corner: 360° in total.
- Two angles of a triangle are 55° and 82°. Find the third.
- Two angles of a triangle are 26° and 98°. Find the third.
- One acute angle of a right triangle is 28°. Find the other acute angle.
- In triangle ABC, A = 40° and B = 65°. What is the exterior angle at C?
- An exterior angle of a triangle is 130°, and one of the two far interior angles is 50°. Find the other far angle.
- The angles of a triangle are x°, (x + 17)° and (3x − 62)°. Find x.
- The angles of a triangle are in the ratio 1 : 2 : 3. How big is the largest?
- Take one exterior angle at each corner of a triangle. What is their sum?
- Each base angle of an isosceles triangle is 52°. Find the angle between the equal sides.
- Why is tearing the corners off ten paper triangles not a proof?
- It checks only those ten triangles, and only roughly
- Paper triangles have more than 180°
- Tearing changes the angles a lot
- A proof needs a protractor
- “The exterior angle at C is 180° − C, so it has nothing to do with A and B.” Both are true at once: 180° − C and A + B are the same number. Using A + B often saves a step, because you skip finding C.
- “A bigger triangle has a bigger angle sum.” Blowing a triangle up makes the sides longer but leaves every angle as it was. The sum is 180° for a triangle on a stamp and for one drawn across a football pitch.
- “The screen adds the three angles to 179.9°, so the rule is wrong.” The trig lab rounds each angle to 0.1°, so the three rounded numbers can miss by 0.1. The exact sum shown on the Sum line is 180°.
- Interior angles of any triangle: A + B + C = 180°.
- Proof: a line through C parallel to AB and two pairs of alternate angles.
- Exterior angle at C = A + B, the two interior angles not next to it.
- One exterior angle per corner adds to 360°.
Sources for this course (7)
- Curriculum: 교육부 고시 제2022-33호 [별책 8] 수학과 교육과정 (ncic.re.kr/inv/org/download.do?year=2022&seq=10003559&orgType=ogi4, PDF read 2026-10-19: 초등학교 3~4학년군 [4수03-08] 이등변삼각형·정삼각형, [4수03-09] 직각삼각형·예각삼각형·둔각삼각형, [4수03-25] 삼각형과 사각형의 내각의 크기의 합을 여러 가지 방법으로 추론; 초등학교 5~6학년군 [6수03-01] 합동·대응점·대응변·대응각, [6수03-14] 삼각형의 넓이(높이가 삼각형의 외부에 있는 것도 다룸); 중학교 1~3학년군 [9수03-03] 삼각형의 작도, [9수03-04] 삼각형의 합동 조건, [9수03-05] 삼각형의 내각과 외각, [9수03-09] 이등변삼각형의 성질을 합동 조건으로 정당화하고 증명이라는 용어 도입, [9수03-12]·[9수03-13] 닮음·닮음비·삼각형의 닮음 조건, [9수03-15] 피타고라스 정리를 추측하고 정당화, 역은 직관적으로 이해하고 직각삼각형 판별에 사용; 용어와 기호 대변, 대각, 내각, 외각, 증명, 닮음비, △ABC, ≡, ∽)
- Textbook: OpenStax, Contemporary Mathematics, 10.3 Triangles (triangle from three noncollinear points; acute, right, obtuse; side a opposite angle A; hypotenuse and legs; angle sum 180° by a parallel line and alternate interior angles; x, x + 17, 3x − 62 gives 45°, 62°, 73°; SSS, SAS, ASA, AAS; similar triangles, scaling factor; tower shadow 180 ft with a 9 ft truck and 15 ft shadow gives 108 ft; Thales and the pyramid: openstax.org/books/contemporary-mathematics/pages/10-3-triangles)
- OpenStax Prealgebra 2e, 9.3 Use Properties of Angles, Triangles, and the Pythagorean Theorem (a² + b² = c²; Pythagoras around 500 BCE; 13 ft ladder 5 ft from the wall; 10 inch bracket about 7.1 inches; similar triangles: openstax.org/books/prealgebra-2e/pages/9-3-use-properties-of-angles-triangles-and-the-pythagorean-theorem)
- 9.4 Use Properties of Rectangles, Triangles, and Trapezoids (A = ½bh from half a rectangle; isosceles and equilateral; equilateral perimeter 93 in: .../pages/9-4-use-properties-of-rectangles-triangles-and-trapezoids)
- OpenStax Precalculus 2e, 8.2 Non-right Triangles: Law of Cosines (Heron’s formula with semi-perimeter s; sides 10, 15, 7 give s = 16 and area about 29.4; Chicago lot 62.4, 43.5, 34.1 m gives about 711.4 m²: openstax.org/books/precalculus-2e/pages/8-2-non-right-triangles-law-of-cosines)
- History for story blocks only: Euclid’s Elements, D. E. Joyce edition, Clark University (Book I Prop. 20 triangle inequality and Heron’s shortest-path problem; Prop. 26 and its guide on congruence theorems, SSA ambiguity and side-side-right angle; Prop. 32 exterior angle and angle sum, Proclus corollaries; Prop. 47 and its guide on Old Babylonians ca. 1900 to 1600 BCE, Plimpton 322, the Zhou bi hypotenuse diagram, Proclus; Prop. 48 converse; Book VI Prop. 4 equiangular triangles: mathcs.clarku.edu/~djoyce/elements/bookI/propI47.html)
- MacTutor History of Mathematics, University of St Andrews, Heron of Alexandria (formula proved in Metrica Book I; eclipse of 13 March 62; side of 720 about 26 5/6: mathshistory.st-andrews.ac.uk/Biographies/Heron/) MacTutor, "Euclid of Alexandria" (born about 325 BC, died about 265 BC; dating for the Elements, about 2,300 years ago: mathshistory.st-andrews.ac.uk/Biographies/Euclid/, checked 2026-10-19).
Checked: 2026-10-19