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A triangle is a polygon with three sides and three interior angles. Triangles can be classified using their side lengths and angle measures.
Concept ### Classification by Sides

Concept ### Classification by Angles

- Scalene triangle: All sides have different length.
- Isosceles triangle: Exactly two sides have the same length.
- Equilateral triangle: All three sides have the same length.

- Acute triangle: All angles are acute, or measure less than $90^\circ.$
- Obtuse triangle: Exactly one angle is obtuse, or measures between $90^\circ$ and $180^\circ,$ exclusive.
- Right triangle: One angle is a right angle, or measures $90^\circ.$
- Equiangular triangle: All angles are congruent and measure $60^\circ.$

If two sides of a triangle are congruent, then the angles opposite them are congruent.

Based on the diagram above, the following relation holds true.

$\overline{NM}\cong \overline{NK}$ $\ \Rightarrow \$ $\angle M\cong \angle K$

The Isosceles Triangle Theorem is also known as the **Base Angles Theorem**.

This theorem will be proven using congruent triangles. Consider $\triangle MNK,$ a triangle with two congruent sides.

In this triangle, let $P$ be the point of intersection of the angle bisector of $\angle N$ and $\overline{MK}.$

From the diagram, the following features of $\triangle MNP$ and $\triangle KNP$ can be observed.

Feature | Reasoning |
---|---|

$\angle MNP \ \cong \ \angle KNP$ | Definition of an angle bisector. |

$\overline{MN} \ \cong \ \overline{KN}$ | Given. |

$\overline{NP} \ \cong \ \overline{NP}$ | Reflexive Property of Congruence. |

Therefore, $\triangle MNP$ and $\triangle KNP$ have two pairs of corresponding congruent sides and one pair of included congruent angles. By the Side-Angle-Side Congruence Theorem, $\triangle MNK$ and $\triangle KNP$ are congruent triangles. $\begin{aligned} \triangle MNK \cong \triangle KNP \end{aligned}$ Corresponding parts of congruent figures are congruent. Therefore, $\angle M$ and $\angle K$ are congruent.

$\angle M \cong \angle K$

It has been proven that if two sides in a triangle are congruent, the angles opposite them are congruent.

Classify the triangle by its sides and its angles.

Show Solution

We can begin by classifying the triangle by its side lengths. First, we must use the distance formula to calculate the length of the sides. We'll start with $\overline{AB}.$
Thus, the side $\overline{AB}$ is $6$ units long. The lengths of the other sides can be calculated in the same way.

$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }$

SubstitutePointsSubstitute $\left({\color{#0000FF}{3.6,4.8}}\right)$ & $\left({\color{#009600}{0,0}}\right)$

$AB = \sqrt{\left({\color{#0000FF}{3.6}}-{\color{#009600}{0}}\right)^2 + \left({\color{#0000FF}{4.8}}-{\color{#009600}{0}}\right)^2}$

SimpTermsSimplify terms

$AB=\sqrt{3.6^2+4.8^2}$

CalcPowCalculate power

$AB=\sqrt{12.96+23.04}$

AddTermsAdd terms

$AB=\sqrt{36}$

CalcRootCalculate root

$AB=6$

Side | Distance formula | Length |
---|---|---|

$\overline{AB}$ | $\sqrt{(3.6-0)^2+(4.8-0)^2}$ | $6$ |

$\overline{BC}$ | $\sqrt{(3.6-5)^2+(4.8-0)^2}$ | $5$ |

$\overline{AC}$ | $\sqrt{(5-0)^2+(0-0)^2}$ | $5$ |

We have found that two of the sides, $\overline{BC}$ and $\overline{AC},$ have the same length. Thus, it is an isosceles triangle. Consequently, the base angles, $\angle A$ and $\angle B,$ are congruent and acute. Since the third angle, $\angle C,$ is also acute we can conclude that it is an acute triangle. Thus, the triangle can be classified as isosceles and acute.

A triangle contains three *interior* angles and creates three *exterior* angles.

Suppose one side of a triangle is extended. The angle created is called an exterior angle.

Because a triangle has three sides, three exterior angles exist. Each exterior angle and its corresponding interior angle are supplementary.

$m\angle A + m\angle A' = 180 ^\circ$

At each vertex, the exterior angle can be defined in two ways.

Consider $\triangle ABC.$

The sum of the interior angles of $\triangle ABC$ is $180^\circ.$

$m\angle A+m\angle B + m\angle C=180^\circ$

To begin, draw a line, $\overleftrightarrow{PQ},$ that passes through $B$ and is parallel to $\overline{AB}.$ $\overleftrightarrow{PQ}$ and $\triangle ABC$ create three angles $\angle PBA,$ $\angle ABC,$ and $\angle CBQ.$

Together, the three angles make a straight angle. Thus, the sum of their measures is $180^\circ.$ $m\angle PBA+m\angle ABC+m\angle CBQ=180^\circ$ Because, $\overleftrightarrow{PQ} \parallel \overline{AC}$ and $\overline{AB}$ is a transversal, $\angle PBA$ and $\angle A$ are alternate interior angles. Thus, according to the Alternate Interior Angles Theorem, $\angle PBA \cong \angle A.$ By the same reasoning, $\angle CBQ \cong \angle C.$

Two congruent angles have the same measure. This can be used to rewrite the sum of the three angles. $m\angle A+m\angle ABC+m\angle C=180^\circ$ Therefore, the sum of the interior angles of a triangle is $180 ^\circ.$ This can be summarized in a flowchart proof.

Find the measures of the angles that are marked in the figure.

Show Solution

To begin, we'll determine the unknown interior angle measures. The Interior Angles Theorem states that the sum of the measures of the interior angles is $180^\circ.$ We can use this to find $x,$ and in turn to find $m\angle A$ and $m\angle B.$
Since $x=58^\circ,$ $m\angle A=58 ^\circ.$ We can substitute $x=58^\circ$ into the given expression for $\angle B.$
$m\angle B=1.5x+6^\circ = 1.5\cdot 58^\circ + 6^\circ = 93^\circ$
Lastly, we can find the measure of the exterior angle, $\angle y.$ Since this angle is supplementary to $\angle A,$ the sum of their measures is $180 ^\circ.$
The measures of the angles marked in the figure are shown below. $\begin{aligned}
&m\angle A=58^\circ \\ &m\angle B=93^\circ \\
&m\angle y=122 ^\circ \end{aligned}$

${\color{#0000FF}{m\angle A}} + {\color{#009600}{m\angle B}} + {\color{#FF0000}{m\angle C}} = 180^\circ$

SubstituteValuesSubstitute values

${\color{#0000FF}{x}} + {\color{#009600}{1.5x+6^\circ}} + {\color{#FF0000}{29^\circ}} = 180^\circ$

SimpTermsSimplify terms

$2.5x+35^\circ=180^\circ$

SubEqn$\text{LHS}-35^\circ=\text{RHS}-35^\circ$

$2.5x=145^\circ$

DivEqn$\left.\text{LHS}\middle/2.5\right.=\left.\text{RHS}\middle/2.5\right.$

$x=58^\circ$

$m\angle A + m\angle y =180 ^\circ$

Substitute$m\angle A={\color{#0000FF}{58^\circ}}$

${\color{#0000FF}{58^\circ}} + m\angle y =180 ^\circ$

SubEqn$\text{LHS}-58^\circ=\text{RHS}-58^\circ$

$m\angle y =122 ^\circ$

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