We have three rods measuring 2, 3 and 4 m. Can their endpoints be joined into a triangular frame? What if the longest rod measures 5 m? That one-metre difference separates a real triangle from a flattened arrangement along one line.
A triangle consists of three noncollinear points and the three segments joining them. Each side is shorter than the sum of the other two, and its interior angles sum to 180° in a Euclidean plane. We will use parallel lines and transversal angles.
Will the three sides close?
Let a,b,c be positive side lengths. A nondegenerate triangle requires:
a + b > c, a + c > b, b + c > a.
Two sides make a broken path between the third side’s endpoints; the straight path is shorter. The inequalities are therefore strict. If the longest side is known, checking it is enough: the other two inequalities hold automatically for positive lengths. These conditions are also sufficient for a triangle to exist: see the commentary on construction I.22. Euclid I.20.
For 2,3,4 we have 2+3>4, so the frame is possible. For 2,3,5 there is equality: the shorter rods only fit along the longest on a straight line. This is a degenerate case, not a triangle under our definition. If the longest exceeds 5 m, they cannot close at all. This checks geometric shape, not material strength.
2 + 3 > 4: a triangle exists
2 + 3 = 5: degeneracy
If two sides measure 2 and 3 m, the third must be less than 5 and also greater than their difference 3−2=1 m:
1 < c < 5.
The triangle flattens at both boundaries. Its perimeter is the sum of the three lengths: 9 m for the 2,3,4 frame.
Where 180° comes from
Call the interior angles at A,B,C α,β,γ. Draw a line through C parallel to AB. The alternate-interior-angle property reproduces α and β on it. Together with γ they fill a straight angle:
α + β + γ = 180°.
This explains the sum for every plane triangle without using a protractor. For example, in a different triangle, two angles of 40° and 65° leave 180°−40°−65°=75° for the third. Euclid I.32.
An exterior angle at a vertex is formed by extending a side beyond that vertex. It forms a linear pair with the interior angle, so it equals 180° minus that angle: the sum of the other two interior angles. With an interior angle of 75°, the exterior angle is 105°=40°+65°.
The proof uses parallelism in a Euclidean plane. The 180° sum cannot simply be transferred to a spherical triangle bounded by great-circle arcs. The limit of the plane model.
Two different classifications
By its sides a triangle is scalene if all lengths differ, isosceles if at least two are equal, and equilateral if all three are equal. We use the inclusive definition: an equilateral triangle is a special isosceles triangle.
By its angles it is acute if all angles are acute, right if one is 90°, and obtuse if one exceeds 90°. Two right or two obtuse angles are impossible: they would use up or exceed 180°, leaving no positive third angle.
Both classifications apply at once: a right triangle may be scalene or isosceles. “Scalene” alone does not determine the largest angle.
A larger side faces a larger angle
Within one triangle a larger side lies opposite a larger angle, and conversely. The angle opposite AB is at C, not A or B. In the 2,3,4 frame, the largest angle therefore faces the 4 m side. This does not yet give its numerical measure. Euclid I.19.
Before assembling the frame, check the strict side inequalities. The angle sum and the larger-side/larger-opposite-angle relationship can then reveal a contradictory sketch.