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IntermediateGrade 725 min21 / 82

Triangles

Classifying triangles by angles and by sides, a proof that the angles add up to 180°, the exterior angle theorem, the angles of isosceles and equilateral triangles, the triangle inequality, the larger side opposite the larger angle, and the median, angle bisector, altitude and midline — with many angle-chasing problems.

Check yourself
In this lesson you will learn
  • Classify triangles by their angles and by their sides
  • Find unknown angles, including those of isosceles triangles, with the angle sum and the exterior angle theorem
  • Use the triangle inequality to check whether a triangle exists, and compare sides by the opposite angles
  • Draw the median, angle bisector, altitude and midline, and solve angle problems with them

Aysel cut a triangle of any shape out of paper, tore off its three corners and put them side by side with the vertices at one point. The corners lined up along a straight line — a straight angle. Murad repeated the experiment with a completely different, long and thin triangle, and got the same result. This is no accident: the angles of every triangle add up to 180°. In this lesson we prove it with the angles from the lesson “Angles and parallel lines”, and then study how the sides, angles and special segments of a triangle are related.

The triangle and its types

Definition
Triangle

A figure made of three points that do not lie on one line and the three segments joining them in pairs. The points are the vertices and the segments are the sides; we write △ABC. A side is often named by the small letter of the opposite vertex: a = BC (opposite A), b = AC, c = AB.

By anglesBy sides
Acute: all angles are acuteScalene: all sides are different
Right: one angle is a right angleIsosceles: two sides are equal
Obtuse: one angle is obtuseEquilateral: all three sides are equal

The two classifications work together: a triangle with angles 90°, 45°, 45°, for example, is both right and isosceles. An equilateral triangle is a special case of an isosceles one. In a right triangle the sides of the right angle are the legs, and the side opposite it is the hypotenuse.

The angle sum and the exterior angle

∠A + ∠B + ∠C = 180°
where:
  • ∠A, ∠B, ∠Cthe interior angles of the triangle

The angle sum theorem: the interior angles of any triangle add up to 180°.

12ACBa
∠1 = ∠A and ∠2 = ∠C (alternate angles), and the three angles at B together form a straight angle.

Proof. Through vertex B draw the line a parallel to AC (by the parallel postulate there is exactly one). At B three angles lie side by side and form a straight angle: ∠1 + ∠B + ∠2 = 180°. Since a ∥ AC and AB is a transversal, ∠1 = ∠A (alternate angles). In the same way, with the transversal BC, ∠2 = ∠C. Substituting gives ∠A + ∠B + ∠C = 180°. This is exactly what Aysel saw on paper.

  • A triangle has at most one right or obtuse angle, so at least two of its angles are acute.
  • The acute angles of a right triangle add up to 90°: ∠A + ∠B = 180° − 90°.
  • Each angle of an equilateral triangle is 180° ÷ 3 = 60°.
  • If two angles of one triangle equal two angles of another, the third angles are equal too.
The third angle

1) Two angles of a triangle are 48° and 67°. Find the third angle and say what type of triangle it is.
2) One acute angle of a right triangle is 34°. Find the other acute angle.
3) The angles of a triangle are in the ratio 2 : 3 : 4. Find them.

Show solution
1) ∠C = 180° − (48° + 67°) = 180° − 115° = 65°. All three angles are less than 90°, so the triangle is acute.
2) The acute angles add up to 90°: 90° − 34° = 56°.
3) Let the angles be 2k, 3k, 4k: 9k = 180°, k = 20°.
Answer: 40°, 60°, 80° — an acute triangle.
Setting up an equation

1) In triangle ABC, ∠B is 20° larger than ∠A, and ∠C is 3 times ∠A. Find the angles and the type of the triangle.
2) Is there a triangle with angles 100°, 50° and 40°?

Show solution
1) Let ∠A = x: x + (x + 20°) + 3x = 180°, 5x = 160°, x = 32°.
Answer: ∠A = 32°, ∠B = 52°, ∠C = 96°. 96° > 90°, so the triangle is obtuse.
2) 100° + 50° + 40° = 190° ≠ 180°, so no such triangle exists.
Definition
Exterior angle of a triangle

An angle adjacent to an interior angle of the triangle; you get it by extending one side beyond a vertex. At each vertex, exterior angle = 180° − interior angle.

ABCD
Exterior angle: ∠BCD = ∠A + ∠B.
∠BCD = ∠A + ∠B
where:
  • ∠BCDthe exterior angle at C (D lies on AC extended beyond C)
  • ∠A, ∠Bthe two interior angles not adjacent to it

Exterior angle theorem: an exterior angle of a triangle equals the sum of the two interior angles not adjacent to it. Proof: ∠BCD = 180° − ∠C, and ∠A + ∠B is also 180° − ∠C. Consequence: an exterior angle is larger than each interior angle not adjacent to it.

Problems with the exterior angle

1) In triangle ABC the exterior angle at C is 130° and ∠A = 55°. Find ∠B.
2) An exterior angle of a triangle is 120°, and the two interior angles not adjacent to it are in the ratio 1 : 3. Find the angles of the triangle.
3) The angles of a triangle are 50°, 60° and 70°. Take one exterior angle at each vertex and find them. What is their sum?

Show solution
1) ∠A + ∠B = 130°, ∠B = 130° − 55° = 75°. Check: ∠C = 180° − 130° = 50°, 55° + 75° + 50° = 180°.
2) Let the interior angles be x and 3x: x + 3x = 120°, x = 30°. The angles are 30° and 90°, and the third angle is 180° − 120° = 60° — a right triangle.
3) Exterior angles: 180° − 50° = 130°, 180° − 60° = 120°, 180° − 70° = 110°. Sum: 130° + 120° + 110° = 360°. This is no accident: the three pairs of adjacent angles add up to 3 · 180° = 540°, and subtracting the 180° of the interior angles leaves 360°.

Isosceles and equilateral triangles

Definition
Isosceles triangle

A triangle with two equal sides. The equal sides are the legs, the third side is the base. The angle between the legs is the apex angle, and the two angles at the base are the base angles.

The main property of an isosceles triangle: the base angles are equal. The converse is also true: if two angles of a triangle are equal, the sides opposite them are equal, so the triangle is isosceles. These facts will be proved in the lesson “Congruent triangles”; for now we use them to calculate angles. In an equilateral triangle every pair of sides is equal, so all three angles are equal, 60° each; conversely, a triangle with three 60° angles is equilateral.

β + 2α = 180° ⇒ α = (180° − β) / 2, β = 180° − 2αβ + 2α = 180° ⇒ α = (180° − β) / 2, β = 180° − 2α
where:
  • βthe apex angle (between the legs)
  • αa base angle (each of the two)

The angle sum for an isosceles triangle: the apex angle plus the two equal base angles make 180°. Hence the base angles are always acute: α < 90°.

Angles of an isosceles triangle

1) The apex angle of an isosceles triangle is 40°. Find the base angles.
2) A base angle is 35°. Find the apex angle and the type of the triangle.
3) In isosceles triangle ABC (AB = BC) the exterior angle at B is 100°. Find the angles of the triangle.
4) A base angle is twice the apex angle. Find the angles.

Show solution
1) α = (180° − 40°) / 2 = 70°. Check: 40° + 70° + 70° = 180°.
2) β = 180° − 2 · 35° = 110° > 90°, so the triangle is obtuse.
3) AB = BC, so AC is the base and ∠A = ∠C. The exterior angle at B equals ∠A + ∠C = 100°, so ∠A = ∠C = 50° and ∠B = 180° − 100° = 80°.
4) Let β = x and α = 2x: x + 2x + 2x = 180°, x = 36°.
Answer: 36°, 72°, 72°.
An angle is given — but which one?

One angle of an isosceles triangle is a) 100°; b) 50°. Find the other two angles.

Show solution
a) 100° cannot be a base angle: two such angles would already make 200° > 180°. So 100° is the apex angle, and the base angles are (180° − 100°) / 2 = 40° and 40°.
b) Two cases are possible.
• If 50° is the apex angle, the base angles are (180° − 50°) / 2 = 65° and 65°.
• If 50° is a base angle, the other base angle is also 50° and the apex angle is 180° − 100° = 80°.
Answer: 65° and 65°, or 50° and 80°.

Sides and angles: the triangle inequality

The shortest path between two points is the segment joining them. So going straight from A to C is always shorter than going through B: AC < AB + BC. This holds for every side of a triangle.

a < b + c, b < a + c, c < a + b
where:
  • a, b, cthe sides of the triangle

Triangle inequality: each side of a triangle is shorter than the sum of the other two. Another form follows: each side is longer than the difference of the other two, |b − c| < a < b + c.

Does the triangle exist?

1) Is there a triangle with sides a) 3 cm, 4 cm, 8 cm; b) 5 cm, 6 cm, 7 cm; c) 2 cm, 3 cm, 5 cm?
2) Two sides of a triangle are 3 cm and 8 cm. Between which values is the third side? If its length is a whole number of centimetres, how many such triangles are there?
3) Two sides of an isosceles triangle are 4 cm and 9 cm. Find its perimeter.

Show solution
1) We check the longest side.
a) 3 + 4 = 7 < 8 — no triangle: the short sticks cannot reach both ends of the long one.
b) 5 + 6 = 11 > 7 — the triangle exists.
c) 2 + 3 = 5 — equality: the three points lie on one line, so there is no triangle.
2) 8 − 3 < x < 8 + 3, i.e. 5 cm < x < 11 cm. Whole values: 6, 7, 8, 9, 10 — 5 triangles.
3) If the legs were 4 cm, then 4 + 4 = 8 < 9 — no such triangle. So the legs are 9 cm and the base is 4 cm: 9 + 9 > 4.
Answer: P = 9 + 9 + 4 = 22 cm.
a > b ⇔ ∠A > ∠B
where:
  • a, btwo sides of the triangle (a = BC, b = AC)
  • ∠A, ∠Bthe angles opposite these sides

In a triangle the larger angle lies opposite the larger side, and the larger side opposite the larger angle; equal angles lie opposite equal sides.

Why? Let AC > AB. Mark point D on AC with AD = AB. Triangle ABD is isosceles, so ∠ABD = ∠ADB. But ∠ADB is an exterior angle of triangle BDC, so it is larger than ∠C. Then ∠B > ∠ABD = ∠ADB > ∠C: the angle B opposite AC is larger than the angle C opposite AB.

Comparing sides and angles

1) In triangle ABC, AB = 5 cm, BC = 7 cm, AC = 9 cm. Which angle is the largest and which is the smallest?
2) In triangle ABC, ∠A = 50° and ∠B = 60°. List the sides from shortest to longest.
3) Which side of a right triangle is the longest? And of an obtuse triangle?

Show solution
1) ∠B lies opposite the longest side AC, so it is the largest angle. ∠C lies opposite the shortest side AB, so it is the smallest.
2) ∠C = 180° − 50° − 60° = 70°. Since ∠A < ∠B < ∠C, the opposite sides are ordered the same way: BC < AC < AB.
3) The right angle is the largest angle of the triangle (the other two are acute), so the hypotenuse is the longest side. In an obtuse triangle the longest side is the one opposite the obtuse angle.

Median, bisector, altitude and midline

Three kinds of special segments come out of the vertices of a triangle; they are often drawn as helper lines in problems. Every triangle has three medians, three angle bisectors and three altitudes.

Definition
Median

The segment joining a vertex of a triangle to the midpoint of the opposite side.

Definition
Angle bisector of a triangle

The part of the bisector of an angle of the triangle from the vertex to the opposite side.

Definition
Altitude

The perpendicular dropped from a vertex of the triangle to the line containing the opposite side.

Definition
Midline

The segment joining the midpoints of two sides of a triangle. Every triangle has three midlines.

ABCMBM — medianABCLBL — angle bisectorABCHBH — altitudeABCHBH — altitude outsidethe triangleABCHCH — altitudeto the hypotenuseABCMNMN — midline
A median halves a side, a bisector halves an angle, and an altitude is perpendicular to the opposite side (or its extension).

In an acute triangle all three altitudes lie inside. In a right triangle two altitudes coincide with the legs. In an obtuse triangle the altitudes from the acute-angled vertices fall outside the triangle — to draw them you extend the opposite side. In an isosceles triangle the bisector drawn from the apex to the base is also a median and an altitude; in a scalene triangle the median, bisector and altitude from one vertex are three different segments. The key property of the midline — it is parallel to the third side and half as long — will be proved in the lesson “Quadrilaterals: parallelogram, rectangle, rhombus, square and trapezoid”.

∠AOB = 90° + ∠C / 2∠AOB = 90° + ∠C / 2
where:
  • Othe point where the bisectors of angles A and B meet
  • ∠Cthe third angle of the triangle

The angle between two bisectors. Proof: in triangle AOB, ∠OAB + ∠OBA = (∠A + ∠B) / 2 = (180° − ∠C) / 2 = 90° − ∠C / 2, so ∠AOB = 180° − (90° − ∠C / 2) = 90° + ∠C / 2.

The angle between bisectors

1) In triangle ABC, ∠A = 70° and ∠B = 50°. The bisectors of angles A and B meet at O. Find ∠AOB.
2) In triangle ABC the bisectors of angles A and B meet at an angle of 125°. Find ∠C.

Show solution
1) ∠C = 180° − 70° − 50° = 60°, so ∠AOB = 90° + 60° / 2 = 120°. Check: in triangle AOB, 180° − 35° − 25° = 120°.
2) Use the formula in reverse: 90° + ∠C / 2 = 125°, ∠C / 2 = 35°, ∠C = 70°.

This plan helps with angle-chasing problems:

  1. 1
    Mark the figure

    Write the known angles, equal sides, right angles and the equal halves made by bisectors on the drawing.

  2. 2
    Look for connections

    A triangle sums to 180°, adjacent angles on a line sum to 180°, vertical angles are equal, parallel lines give Z, F and C, an isosceles triangle has equal base angles, and an exterior angle is the sum of the two far angles.

  3. 3
    Call the unknown x

    If the angles are linked (“twice as large”, “20° smaller”), call one of them x and set up an equation.

  4. 4
    Check

    In every triangle the angles must add up to 180°, every angle must be positive, and the larger angle must lie opposite the larger side.

An altitude and a bisector from one vertex

1) In triangle ABC, ∠B = 70° and ∠C = 30°. The altitude AH and the bisector AL are drawn from A. Find ∠HAL.
2) In triangle ABC, ∠C = 90°, ∠A = 35°, and CH is an altitude. Find ∠ACH and ∠BCH.

Show solution
1) ∠A = 180° − 70° − 30° = 80°, and the bisector halves it: ∠BAL = 40°. In the right triangle ABH, ∠BAH = 90° − 70° = 20°. Point H lies between B and L, so ∠HAL = ∠BAL − ∠BAH = 40° − 20° = 20°.
2) In triangle ACH, ∠AHC = 90°, so ∠ACH = 90° − 35° = 55°. Then ∠BCH = 90° − 55° = 35° — equal to ∠A.
A median and perimeters

In triangle ABC the median BM is drawn, AB = 7 cm and BC = 10 cm. Find the difference between the perimeters of triangles CBM and ABM.

Show solution
M is the midpoint of AC, so AM = MC, and BM belongs to both triangles.
P(CBM) − P(ABM) = (BC + BM + MC) − (AB + BM + AM) = BC − AB = 10 − 7 = 3 cm.
Notice that neither AC nor BM was needed.
Check yourself: fill in the gap
  1. 1.Two angles of a triangle are 70° and 45°. The third angle is °.
  2. 2.The apex angle of an isosceles triangle is 80°. Each base angle is °.
  3. 3.An exterior angle is 140°, and one of the interior angles not adjacent to it is 60°. The other one is °.
  4. 4.Two sides of a triangle are 1 cm and 6 cm, and the third side is a whole number. It is cm.
  5. 5.One acute angle of a right triangle is 27°. The other is °.

Key points

  • By angles triangles are acute, right or obtuse; by sides they are scalene, isosceles or equilateral.
  • The angles add up to 180° (proved with a line through a vertex parallel to the opposite side); an exterior angle equals the sum of the two non-adjacent interior angles.
  • In an isosceles triangle the base angles are equal: α = (180° − β) / 2; all angles of an equilateral triangle are 60°.
  • Each side is shorter than the sum of the other two — checking the longest side is enough; the larger angle lies opposite the larger side.
  • A median halves the opposite side, a bisector halves the angle, an altitude is perpendicular to the opposite side, a midline joins the midpoints of two sides; the angle between two bisectors is 90° + ∠C / 2.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
What is an exterior angle of a triangle equal to?