# Triangles

**Triangles** are polygons (shapes) with three sides and three angles, which can be formed by connecting any three points in a plane. They are one of the first shapes studied in geometry.

Triangles are particularly important because arbitrary polygons (with 4, 5, 6, or \(n\) sides) can be decomposed into triangles. Thus, understanding the basic properties of triangles allows for deeper study of these larger polygons as well. Interestingly, the triangle is the only rigid polygon formed out of straight line segments, meaning that if the three lengths of the sides are given, the measurements correspond to a unique triangle. Because of this, it is often possible, given some information about a triangle (e.g. some side lengths and some angles), to determine additional facts about the triangle.

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## Sum of Angles in a Triangle

If \(ABC\) is a triangle, then \(\angle ABC + \angle BCA + \angle CAB = 180^\circ.\) In other words, the sum of the \(3\) internal angles in any triangle is \(180^\circ.\)

Draw line \(DE\) that is parallel to \(AB\) and passes through \(C\) as in the above figure.

Since \( DE \parallel AB \), applying the principle of alternate interior angles shows that \( \angle DCA = \angle CAB \) and \( \angle BCE = \angle CBA.\)

Since angles in a line sum up to 180 degrees, \( \angle DCA + \angle ACB + \angle BCE = 180^ \circ \).

Thus, we conclude that \( \angle CAB + \angle ACB + \angle CBA = \angle DCA + \angle ACB + \angle BCE = 180^ \circ \). \(_\square\)

## Triangle Inequality

Main article: Triangle Inequality

Triangles have the property that the sum of any two sides of the triangle is always strictly greater than the third side. This property, known as the **triangle inequality**, is explored in the wiki linked above.

## Classifying Triangles

Main article: Classification of Triangles

Triangles can be classified into different categories based on their sides and angles. For example, a triangle with one angle of measure \(90^\circ\) is known as a right triangle, while a triangle with sides of all equal length is known as an equilateral triangle. These classifications and many others are explored in the wiki linked above.

## Area of Triangles

Main article: Area of a Triangle

When determining the area of a triangle, note that a triangle can be thought of as half of a parallelogram. The following picture should make this point clear:

Because the area of a parallelogram is equal to the product of its base and height, the area of a triangle is simply half of that area.

The area of a triangle is \(A = \frac{bh}{2}\), where \(b\) is the length of the base and \(h\) is the height.

What is the area of a triangle with base 10 and height 6?

The area of this triangle is \(\frac{10 \times 6}{2} = 30.\) \(_\square\)

Nihar and Andrew are trying to find the area of \(\triangle ABC\) using the formula

\[\dfrac{1}{2} \times \text{base} \times \text{height}. \]

Nihar mistakenly multiplies base \(AB\) by the height from \(A\) \((\)instead of \(C).\) He gets a value of 14.

Andrew mistakenly multiplies base \(BC\) by the height from \(C\) \((\)instead of \(A).\) He gets a value of 56.

Find the actual area of triangle \(\triangle ABC\).

For more advanced methods of finding the area of a triangle, such as Heron's formula, see the wiki linked at the top of this section.

## Exterior Angles of Triangles

The measure of an **exterior angle** of a triangle is the sum of its two remote interior angles. **Remote interior angles** are the interior angles of a triangle that are opposite to the exterior angle under consideration.

For **regular polygons**, the formula to find the exterior angle of a polygon is \(\frac{360^\circ}{n},\) where \(n\) is the number of sides.

Prove that \(\angle ABC+\angle CAB=\angle DCA. \)

From the property of triangles-angle sum, the sum of measures of all angles in a triangle is \(180^\circ\). So, in \(\triangle ABC\), it follows that \( \angle ABC + \angle BCA + \angle CAB = 180^\circ \).Since \( \angle BCA \) and \( \angle DCA \) form a straight line, \( \angle BCA + \angle DCA = 180^\circ, \) It follows that both sets of angles are equal to \( 180^\circ: \)

\[ \angle ABC + \angle BCA + \angle CAB = \angle BCA + \angle DCA.\]

Therefore,

\[ \angle ABC + \angle CAB = \angle DCA. \ _\square \]

## Problem Solving with Triangles

## See Also

Interested in learning more about triangles? Check out these pages: