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Describing Triangles


Types of Triangles

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

Classification by Sides


Classification by Angles


Isosceles Triangle Theorem

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

An isosceles triangle with two congruent sides and base angles

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

NMNK\overline{NM}\cong \overline{NK}   \ \Rightarrow \ MK\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 MNK,\triangle MNK, a triangle with two congruent sides.

An isosceles triangles with two congruent sides

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

An isosceles triangle with two congruent sides and an angle bisector

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

Feature Reasoning
MNP  KNP\angle MNP \ \cong \ \angle KNP Definition of an angle bisector.
MN  KN\overline{MN} \ \cong \ \overline{KN} Given.
NP  NP\overline{NP} \ \cong \ \overline{NP} Reflexive Property of Congruence.

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

MK\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 AB.\overline{AB}.
AB=(x2x1)2+(y2y1)2AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }
AB=(3.60)2+(4.80)2AB = \sqrt{\left({\color{#0000FF}{3.6}}-{\color{#009600}{0}}\right)^2 + \left({\color{#0000FF}{4.8}}-{\color{#009600}{0}}\right)^2}
Thus, the side AB\overline{AB} is 66 units long. The lengths of the other sides can be calculated in the same way.
Side Distance formula Length
AB\overline{AB} (3.60)2+(4.80)2\sqrt{(3.6-0)^2+(4.8-0)^2} 66
BC\overline{BC} (3.65)2+(4.80)2\sqrt{(3.6-5)^2+(4.8-0)^2} 55
AC\overline{AC} (50)2+(00)2\sqrt{(5-0)^2+(0-0)^2} 55

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


Angles of a Triangle

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

Interior Angles of a Triangle

The interior angles are the angles on the inside of the triangle.


Exterior Angles of a Triangle

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.

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

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


Interior Angles Theorem

Consider ABC.\triangle ABC.

The sum of the interior angles of ABC\triangle ABC is 180.180^\circ.

mA+mB+mC=180m\angle A+m\angle B + m\angle C=180^\circ


Interior Angles Theorem

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

Together, the three angles make a straight angle. Thus, the sum of their measures is 180.180^\circ. mPBA+mABC+mCBQ=180 m\angle PBA+m\angle ABC+m\angle CBQ=180^\circ Because, PQAC\overleftrightarrow{PQ} \parallel \overline{AC} and AB\overline{AB} is a transversal, PBA\angle PBA and A\angle A are alternate interior angles. Thus, according to the Alternate Interior Angles Theorem, PBAA. \angle PBA \cong \angle A. By the same reasoning, CBQC.\angle CBQ \cong \angle C.

Two congruent angles have the same measure. This can be used to rewrite the sum of the three angles. mA+mABC+mC=180 m\angle A+m\angle ABC+m\angle C=180^\circ Therefore, the sum of the interior angles of a triangle is 180.180 ^\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 180.180^\circ. We can use this to find x,x, and in turn to find mAm\angle A and mB.m\angle B.
mA+mB+mC=180{\color{#0000FF}{m\angle A}} + {\color{#009600}{m\angle B}} + {\color{#FF0000}{m\angle C}} = 180^\circ
x+1.5x+6+29=180{\color{#0000FF}{x}} + {\color{#009600}{1.5x+6^\circ}} + {\color{#FF0000}{29^\circ}} = 180^\circ
Since x=58,x=58^\circ, mA=58.m\angle A=58 ^\circ. We can substitute x=58x=58^\circ into the given expression for B.\angle B. mB=1.5x+6=1.558+6=93 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, y.\angle y. Since this angle is supplementary to A,\angle A, the sum of their measures is 180.180 ^\circ.
mA+my=180m\angle A + m\angle y =180 ^\circ
58+my=180{\color{#0000FF}{58^\circ}} + m\angle y =180 ^\circ
my=122m\angle y =122 ^\circ
The measures of the angles marked in the figure are shown below. mA=58mB=93my=122\begin{aligned} &m\angle A=58^\circ \\ &m\angle B=93^\circ \\ &m\angle y=122 ^\circ \end{aligned}
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