Mathematics · Plane geometry · Grade 7

Angles of a triangle

If you tear off the three angles and put them side by side, why do you always get a straight angle?

The situation

Tear the three angles off a paper triangle and put their corners together at one point — whatever the triangle, you get a straight angle. The direct road from Tashkent to Bukhara is always shorter than the detour through Samarkand — the three cities on the map form a triangle. Why is the sum of the angles always 180°, why does the larger angle lie opposite the longer side, and from which three segments can a triangle be built?

Topic

The sum of the angles of a triangle, the right-angled triangle, the bisector property, relations between sides and angles, the triangle inequality. Tearing off the angles of a paper triangle and putting them side by side — 180°; the road between three cities — the triangle inequality

Key ideas

Angle sum 180°
Right-angled triangle
Triangle inequality

Formulas

α + β + γ = 180^{\circ}

The sum of the interior angles of a triangle is a straight angle. Proof: a line drawn through a vertex parallel to the base, and alternate angles.

∠ C B D = α + γ

An exterior angle equals the sum of the two interior angles not adjacent to it (and is greater than each of them).

α + β = 90^{\circ}, a = \frac{c}{2} (α = 30^{\circ})

Right-angled triangle: the acute angles add up to $90^{\circ}$ ; the leg opposite $30^{\circ}$ is half the hypotenuse (and conversely).

M K = M L, M K ⟂ O A, M L ⟂ O B

Property of the angle bisector: a point $M$ on the bisector is equidistant from the sides of the angle; the converse is true too — a point equidistant from the sides lies on the bisector.

a < b + c, b < a + c, c < a + b

Triangle inequality: each side is less than the sum of the other two. The larger angle lies opposite the longer side.

Theory

Angle sum and the exterior angle

In every triangle $α + β + γ = 180^{\circ}$ . Proof: through the vertex $C$ we draw a line parallel to $A B$ ; the alternate interior angles made by the parallel lines and the transversals $A C$ and $B C$ are equal, so the three angles at $C$ fill a straight angle. An exterior angle is the angle adjacent to an interior angle: $∠ C B D = 180^{\circ} - β = α + γ$ . In the first step, drag $C$ : the angles change, but the sum is always $180^{\circ}$ .

Right-angled triangle

If one angle is $90^{\circ}$ , the other two are acute and add up to $90^{\circ}$ . The **leg opposite the $30^{\circ}$ angle is half the hypotenuse: if we reflect the triangle in the leg, we get an isosceles** triangle with a $60^{\circ}$ angle at the apex — that is, an equilateral triangle — with side $2 a = c$ . The converse: if a leg is half the hypotenuse, the angle opposite it is $30^{\circ}$ . Congruence criteria for right-angled triangles: two legs; a leg and an acute angle; the hypotenuse and an acute angle; the hypotenuse and a leg.

Property of the angle bisector

Every point on the bisector of an angle is equidistant from the sides of the angle: if $M K ⟂ O A$ and $M L ⟂ O B$ , then $M K = M L$ . Proof: the right-angled triangles $O M K$ and $O M L$ share the hypotenuse $O M$ and have equal acute angles (the bisector) — so they are congruent by the hypotenuse and an acute angle. The converse: a point inside an angle that is equidistant from the sides lies on the bisector. In the fourth step, drag $M$ : when you leave the bisector, the distances differ.

The relation between sides and angles

In a triangle the larger angle lies opposite the longer side, and conversely. Proof: let $A C > A B$ ; on $A C$ we take the point D with $A D = A B$ . Triangle $A B D$ is isosceles: $∠ A B D = ∠ A D B$ . $∠ A D B$ is an exterior angle of triangle $B D C$ , so $∠ A D B > ∠ C$ ; and $∠ A B C > ∠ A B D$ . Thus $∠ B > ∠ C$ . Consequence: in a right-angled triangle the hypotenuse is the longest side; equal sides lie opposite equal angles.

Triangle inequality

Each side is less than the sum of the other two: $c < a + b$ . On the map the direct road is always shorter than the detour. To build a triangle from three segments, all three inequalities must hold: in the sixth step the compass arcs meet only then. In short: $| a - b | < c < a + b$ . If two sides are given, the third can be any number in this range.

About the model

Angles are shown to a tenth of a degree and lengths are rounded; the vertices move in steps of 0.5 units. The distances between cities are straight-line (as the crow flies), not the real length of the road; they are roughly calculated from the geographic coordinates of the cities. The paper pieces are assumed not to change the triangle when they are “torn off”.

Did you know?

That the angles of a triangle add up to $180^{\circ}$ was proved in Euclid's “Elements” (about 300 BC) as Proposition 32 of Book I — together with the statement about the exterior angle.
On a sphere the angles of a triangle add up to more than $180^{\circ}$ : on the Earth, in a triangle formed by the equator and two meridians, both angles at the equator are $90^{\circ}$ — so the sum exceeds $180^{\circ}$ by the angle at the pole.

Inside the experiment

The topic is explained step by step with animation, from simple to advanced: from a real-life picture to the formula and the proof. Then you drag the points yourself to check. Inside are missions, a quiz and a guided tour.