Triangles

Take three points that do not lie on one line and join them with segments. The figure you get is a triangle. The three points are its vertices, the three segments are its sides, and the three angles at the vertices are its angles.

Standard notation. In triangle $ABC$ each side gets the small letter of the vertex opposite it:

$$a=BC,\qquad b=CA,\qquad c=AB.$$

The angles at $A,$ $B$ and $C$ are $\alpha,$ $\beta$ and $\gamma.$ So the side $a$ lies opposite the angle $\alpha,$ the side $b$ opposite $\beta,$ and the side $c$ opposite $\gamma.$

On this page: Sum of the angles · Triangle inequality · Types · Equilateral · Isosceles · Right · Perimeter and area · Special lines · Solved problems · Practice

The Sum of the Angles of a Triangle

In every triangle the three angles add up to ${180^\circ}.$

$$\alpha+\beta+\gamma=180^\circ$$

Why? Draw through the vertex $C$ the line $p$ parallel to $AB.$ The angle between $p$ and $CA$ equals $\alpha,$ because these are alternate angles at the parallel lines $p$ and $AB$ cut by $AC.$ In the same way the angle between $p$ and $CB$ equals $\beta.$ Together with $\gamma$ these two angles make a straight angle at $C,$ so ${\alpha+\beta+\gamma=180^\circ}.$

Some consequences:

A triangle has at most one right or obtuse angle. At least two of its angles are acute.
If two angles are known, the third is ${180^\circ}$ minus their sum. For ${\alpha=50^\circ}$ and ${\beta=65^\circ}$ we get ${\gamma=180^\circ-115^\circ=65^\circ}.$
In a right triangle the two acute angles add up to ${90^\circ}.$
Each angle of an equilateral triangle is ${60^\circ}.$

The exterior angle

Extend the side $AC$ beyond $C$ to a point $D.$ The angle ${\delta=\angle BCD}$ between the side $CB$ and this extension is an exterior angle of the triangle at the vertex $C.$ It lies next to the interior angle $\gamma,$ and ${\delta+\gamma=180^\circ}.$

An exterior angle of a triangle equals the sum of the two interior angles that are not next to it: ${\delta=\alpha+\beta}.$

Indeed, ${\delta=180^\circ-\gamma}$ and ${\alpha+\beta=180^\circ-\gamma}.$ So an exterior angle is greater than each of the two interior angles that are not next to it.

The Triangle Inequality

A straight segment is the shortest path between two points. Going from $B$ to $C$ directly is shorter than going through $A,$ so ${a\lt b+c}.$ The same is true for every side.

Each side of a triangle is shorter than the sum of the other two sides and longer than their difference:

$$|b-c|\lt a\lt b+c.$$

Do three lengths form a triangle? Compare the longest of them with the sum of the other two.

$5,$ $7$ and $11.$ Since ${11\lt 5+7=12},$ such a triangle exists.
$3,$ $4$ and $8.$ Since ${8\gt 3+4=7},$ there is no such triangle: the two shorter segments cannot reach each other.
$2,$ $3$ and $5.$ Since ${5=2+3},$ the three points would lie on one line, so this is not a triangle either.

The larger side lies opposite the larger angle

In every triangle the larger of two sides lies opposite the larger angle, and the larger of two angles lies opposite the larger side. Equal sides lie opposite equal angles.

For example, if ${\alpha=40^\circ},$ ${\beta=60^\circ}$ and ${\gamma=80^\circ},$ then ${a\lt b\lt c}.$ In a right triangle the hypotenuse is the longest side, because it lies opposite the right angle, the largest angle of the triangle.

Types of Triangles

By sides a triangle is scalene if all its sides have different lengths, isosceles if two of its sides are equal, and equilateral if all three sides are equal. Equal sides are marked with the same number of small ticks.

By angles a triangle is acute if all its angles are acute, right if one angle is ${90^\circ},$ and obtuse if one angle is greater than ${90^\circ}.$ A small square marks a right angle.

The full name uses both: a triangle with angles ${45^\circ},$ ${45^\circ}$ and ${90^\circ}$ is a right isosceles triangle. How to find the type from the angles or from the side lengths is explained step by step on the page Classification of Triangles.

Equilateral Triangle

A triangle with three equal sides is equilateral. Let its side be $a.$

Its three angles are equal, so each of them is ${180^\circ:3=60^\circ}.$
Each altitude is also a median and an angle bisector. For example, the altitude $CD$ meets $AB$ at its midpoint $D$ and halves the angle at $C.$
The three altitudes are equal. The Pythagorean theorem in the right triangle $ADC$ gives ${h^2=a^2-\left(\frac a2\right)^2=\frac{3a^2}{4}}.$
The orthocenter, the centroid, the center of the inscribed circle and the center of the circumscribed circle coincide (these points are explained below). This point $O$ is the center of the triangle.

For an equilateral triangle with side $a$

$$h=\frac{a\sqrt3}{2},\qquad S=\frac{a^2\sqrt3}{4},\qquad R=\frac{a\sqrt3}{3},\qquad r=\frac{a\sqrt3}{6},$$

where $S$ is the area, $R$ the radius of the circumscribed circle and $r$ the radius of the inscribed circle. Note that ${R=\frac23 h}$ and ${r=\frac13 h}.$

Example. For a side of 6 cm, the altitude is ${h=\frac{6\sqrt3}{2}=3\sqrt3\approx 5.20}$ cm and the area is ${S}={\frac{36\sqrt3}{4}}={9\sqrt3}\approx{15.59}$ cm².

Isosceles Triangle

A triangle with two equal sides is isosceles. The equal sides are its legs, and the third side is its base. The angles at the base are the base angles, and the angle between the legs is the vertex angle. In the figure ${AC=BC}$ and the base is $AB.$

Properties of an isosceles triangle

The base angles are equal: ${\angle A=\angle B}.$ Conversely, a triangle with two equal angles is isosceles.
The altitude to the base is also the median to the base and the bisector of the vertex angle. Its line is the axis of symmetry of the triangle.

These properties are proved with congruent triangles on the pages Congruent Triangles Problems: SAS and ASA, AAS and SSS. The altitudes to the two legs are equal to each other, and so are the medians to the legs and the bisectors of the base angles. But an altitude to a leg is not a median (unless the triangle is equilateral).

If each leg is $b$ and the base is $c,$ the altitude $CD$ halves the base, and the Pythagorean theorem in triangle $ADC$ gives

${h_c=\sqrt{b^2-\frac{c^2}{4}}}.$

Example. The legs are 13 cm and the base is 10 cm. Then ${h_c}={\sqrt{169-25}}={\sqrt{144}}={12}$ cm, and the area is ${\frac12\cdot 10\cdot 12=60}$ cm². If the vertex angle of an isosceles triangle is ${40^\circ},$ each base angle is ${(180^\circ-40^\circ):2=70^\circ}.$

Right Triangle

A triangle with a right angle is a right triangle. The side opposite the right angle is the hypotenuse, and it is the longest side. The two sides of the right angle are the legs.

In the figure ${\gamma=90^\circ},$ the legs are ${a=BC}$ and ${b=CA},$ the hypotenuse is ${c=AB},$ and ${CH=h}$ is the altitude to the hypotenuse. It divides the hypotenuse into the segments ${AH=m}$ and ${HB=n},$ the projections of the legs $b$ and $a$ on the hypotenuse.

$$\alpha+\beta=90^\circ,\qquad c^2=a^2+b^2\quad\text{(the Pythagorean theorem)},$$ $$S=\frac{ab}{2}=\frac{ch}{2},\qquad h=\frac{ab}{c},$$ $$a^2=c\cdot n,\qquad b^2=c\cdot m,\qquad h^2=m\cdot n.$$

Proof of ${a^2=c\cdot n},$ ${b^2=c\cdot m}$ and ${h^2=m\cdot n}$

Triangles $ABC$ and $CBH$ are similar: each has a right angle, and they share the angle $\beta.$ So ${\frac{c}{a}=\frac{a}{n}},$ that is ${a^2=c\cdot n}.$ In the same way triangles $ABC$ and $ACH$ share the angle $\alpha,$ so ${\frac{c}{b}=\frac{b}{m}}$ and ${b^2=c\cdot m}.$

Triangles $ACH$ and $CBH$ are similar too, because ${\angle ACH=90^\circ-\alpha=\beta}.$ So ${\frac{h}{m}=\frac{n}{h}},$ that is ${h^2=m\cdot n}.$ Adding the first two equalities gives the Pythagorean theorem: ${a^2+b^2=c\,(m+n)=c^2}.$ Finally, the area computed from the legs and from the hypotenuse is the same, ${\frac{ab}{2}=\frac{ch}{2}},$ so ${h=\frac{ab}{c}}.$

Trigonometric ratios. For the acute angle $\alpha$

${{\sin\alpha=\frac{a}{c}},\quad{\cos\alpha=\frac{b}{c}},\quad{\tan\alpha=\frac{a}{b}},}$

so ${a=c\sin\alpha=c\cos\beta}$ and ${b=c\sin\beta=c\cos\alpha}.$ More on the page Trigonometric Ratios in a Right Triangle.

The midpoint of the hypotenuse is equally far from the three vertices. It is the center of the circumscribed circle, so ${R=\frac{c}{2}},$ and the median to the hypotenuse is half the hypotenuse: ${m_c=\frac{c}{2}}.$ (Here $m_c$ is the median, not the segment $m.$)
The radius of the inscribed circle is ${r=\frac{a+b-c}{2}}.$
In a triangle with angles ${30^\circ},$ ${60^\circ}$ and ${90^\circ}$ the leg opposite the ${30^\circ}$ angle is half the hypotenuse. In a right isosceles triangle (${45^\circ},$ ${45^\circ},$ ${90^\circ}$) the hypotenuse is ${c=a\sqrt2}.$
The converse of the Pythagorean theorem: if ${c^2=a^2+b^2},$ the angle opposite $c$ is a right angle.

Example. The legs are ${a=15}$ and ${b=20}.$ Then ${c=\sqrt{225+400}=25},$ ${h=\frac{15\cdot 20}{25}=12},$ ${n=\frac{a^2}{c}=\frac{225}{25}=9}$ and ${m=\frac{b^2}{c}=\frac{400}{25}=16}.$ Check: ${m+n=25=c}$ and ${m\cdot n=144=h^2}.$ The circumradius is ${R=12.5}$ and the inradius is ${r=\frac{15+20-25}{2}=5}.$ More problems: Pythagorean Theorem Problems.

Perimeter and Area

The perimeter is the sum of the sides, ${P=a+b+c},$ and half of it is the semiperimeter ${p=\frac{a+b+c}{2}}.$ The area $S$ can be found in several ways; choose the formula that fits the data.

Known	Formula	Example
A side and the altitude to it	${S=\frac12\,a\,h_a}$	${a=10}$ and ${h_a=6}$ give ${S=30}$
Two sides and the angle between them	${S=\frac12\,ab\sin\gamma}$	${a=8},$ ${b=5},$ ${\gamma=30^\circ}$ give ${S=\frac12\cdot 8\cdot 5\cdot\frac12=10}$
The three sides (Heron's formula)	${S}={\sqrt{p(p-a)(p-b)(p-c)}}$	For $13,$ $14,$ $15$ the semiperimeter is ${p=21}$ and ${S=\sqrt{21\cdot 8\cdot 7\cdot 6}=84}$
The radius $r$ of the inscribed circle	${S=p\,r}$	For $13,$ $14,$ $15$ we get ${r=\frac{84}{21}=4}$
The radius $R$ of the circumscribed circle	${S=\frac{abc}{4R}}$	For $13,$ $14,$ $15$ we get ${R}={\frac{13\cdot 14\cdot 15}{4\cdot 84}}={8.125}$

More formulas, also for other figures: Area Formulas.

Altitudes, Medians, Angle Bisectors and Perpendicular Bisectors

An altitude is the perpendicular segment from a vertex to the line containing the opposite side. The three altitudes (or their lines) meet at the orthocenter. See Altitude of a Triangle.
A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, which divides each of them in the ratio ${2:1}$ from the vertex. See Medians of a Triangle.
An angle bisector of a triangle is the segment from a vertex to the opposite side that divides the angle at this vertex into two equal angles. The three bisectors meet at the center of the inscribed circle.
The perpendicular bisector of a side is the line perpendicular to that side through its midpoint. The three perpendicular bisectors meet at the center of the circumscribed circle.
A midsegment joins the midpoints of two sides. It is parallel to the third side and half as long. See Midsegment.

Congruent and Similar Triangles

Two triangles are congruent if one can be moved onto the other so that they coincide; then all their corresponding sides and angles are equal. Three well-chosen equal pairs are enough to prove it: see Congruence of Triangles and the solved problems for SAS and for ASA, AAS and SSS. Triangles with the same shape but possibly different sizes are similar: their angles are equal and their sides are proportional. See Similar Triangles.

Solved Problems

Problem 1. The angles of a triangle are in the ratio ${2:3:4}.$ Find them.

Solution. Let the angles be $2k,$ $3k$ and $4k.$ Then ${2k+3k+4k=180^\circ},$ so ${9k=180^\circ}$ and ${k=20^\circ}.$ The angles are ${40^\circ},$ ${60^\circ}$ and ${80^\circ}.$

Problem 2. Is there a triangle with sides 4 cm, 7 cm and 12 cm?

Solution. No. The longest side is too long: ${12\gt 4+7=11}.$

Problem 3. Two sides of a triangle are 5 cm and 9 cm. What can the third side be? What is it if the triangle is isosceles?

Solution. By the triangle inequality ${9-5\lt x\lt 9+5},$ that is ${4\lt x\lt 14}.$ If the triangle is isosceles, the third side equals one of the other two, so it is 5 cm or 9 cm. Both are between 4 and 14, so both are possible.

Problem 4. The exterior angle at the vertex $C$ is ${130^\circ},$ and ${\alpha=55^\circ}.$ Find $\beta$ and $\gamma.$

Solution. ${\gamma=180^\circ-130^\circ=50^\circ}.$ The exterior angle equals ${\alpha+\beta},$ so ${\beta=130^\circ-55^\circ=75^\circ}.$ Check: ${55^\circ+75^\circ+50^\circ=180^\circ}.$

Problem 5. One angle of an isosceles triangle is ${100^\circ}.$ Find the other two angles.

Solution. The angle of ${100^\circ}$ cannot be a base angle: the two equal base angles would add up to ${200^\circ}.$ So it is the vertex angle, and each base angle is ${(180^\circ-100^\circ):2=40^\circ}.$

Problem 6. The perimeter of an equilateral triangle is 18 cm. Find its altitude and its area.

Solution. The side is ${a=18:3=6}$ cm. Then ${h=\frac{6\sqrt3}{2}=3\sqrt3\approx 5.20}$ cm and ${S}={\frac{36\sqrt3}{4}}={9\sqrt3}\approx{15.59}$ cm².

Problem 7. The legs of a right triangle are 6 cm and 8 cm. Find the hypotenuse, the altitude to the hypotenuse and the segments into which this altitude divides the hypotenuse.

Solution. ${c=\sqrt{36+64}=10}$ cm and ${h=\frac{6\cdot 8}{10}=4.8}$ cm. The projection of the leg 6 is ${\frac{36}{10}=3.6}$ cm, and the projection of the leg 8 is ${\frac{64}{10}=6.4}$ cm. Check: ${3.6+6.4=10}$ and ${3.6\cdot 6.4}={23.04}={4.8^2}.$

Problem 8. Find the area of the triangle with sides 13 cm, 14 cm and 15 cm, and the altitude to the side 14 cm.

Solution. ${p=\frac{13+14+15}{2}=21},$ so by Heron's formula ${S}={\sqrt{21\cdot 8\cdot 7\cdot 6}}={\sqrt{7056}}={84}$ cm². The altitude is ${h=\frac{2S}{14}=\frac{168}{14}=12}$ cm.

Practice Problems

Try them yourself, then open the answer.

1. Two angles of a triangle are ${47^\circ}$ and ${68^\circ}.$ Find the third angle.

${180^\circ-47^\circ-68^\circ=65^\circ}$

2. Is there a triangle with sides 6, 10 and 15? If so, is it acute, right or obtuse?

Yes, because ${15\lt 6+10=16}.$ Since ${15^2}={225}\gt{6^2+10^2}={136},$ the angle opposite the side 15 is obtuse: the triangle is obtuse. (This test is explained on the page Classification of Triangles.)

3. One angle of an isosceles triangle is ${110^\circ}.$ Find the other two angles.

${110^\circ}$ must be the vertex angle, so each base angle is ${(180^\circ-110^\circ):2=35^\circ}.$

4. The exterior angle at the vertex $B$ is ${100^\circ}$ and ${\alpha=30^\circ}.$ Find $\beta$ and $\gamma.$

${\beta=180^\circ-100^\circ=80^\circ}$ and ${\gamma=100^\circ-30^\circ=70^\circ}.$

5. Find the altitude and the area of an equilateral triangle with side 10 cm.

${h=\frac{10\sqrt3}{2}=5\sqrt3\approx 8.66}$ cm and ${S}={\frac{100\sqrt3}{4}}={25\sqrt3}\approx{43.30}$ cm².

6. The legs of a right triangle are 9 and 12. Find the hypotenuse, the area and the altitude to the hypotenuse.

${c=\sqrt{81+144}=15},$ ${S=\frac{9\cdot 12}{2}=54}$ and ${h=\frac{9\cdot 12}{15}=7.2}.$

7. The hypotenuse of a right triangle is 26 and one leg is 10. Find the other leg and the area.

${b}={\sqrt{676-100}}={\sqrt{576}}={24}$ and ${S=\frac{10\cdot 24}{2}=120}.$

8. An isosceles triangle has perimeter 32 cm and base 12 cm. Find the legs, the altitude to the base and the area.

Each leg is ${(32-12):2=10}$ cm. The altitude is ${\sqrt{10^2-6^2}=8}$ cm, and the area is ${\frac12\cdot 12\cdot 8=48}$ cm².

9. Find the area of the triangle with sides 7, 8 and 9.

${p=12},$ so ${S}={\sqrt{12\cdot 5\cdot 4\cdot 3}}={\sqrt{720}}={12\sqrt5}\approx{26.83}.$

10. The altitude to the hypotenuse divides it into segments of 4 and 9. Find the altitude and the legs.

${h=\sqrt{4\cdot 9}=6}.$ The hypotenuse is ${c=13},$ so the legs are ${\sqrt{13\cdot 4}=2\sqrt{13}\approx 7.21}$ and ${\sqrt{13\cdot 9}=3\sqrt{13}\approx 10.82}.$

More about triangles: Classification · Altitude · Median · Congruence · Similar triangles