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Because the angles of a triangle add up to 180°, at least two
of them must be acute (less than 90°). In an **acute
triangle** all angles are acute. A **right triangle** has one right
angle, and an **obtuse triangle** has one obtuse angle.

The **altitude** corresponding to a side is the perpendicular dropped
to the line containing that side from the opposite vertex. The
**bisector** of a vertex is the line that divides the angle at that
vertex into two equal parts. The **median** is the segment joining a
vertex to the midpoint of the opposite side.
See Figure 1.

**Figure 1:** Notations for an arbitrary triangle of sides *a*, *b*, *c*
and vertices *A*, *B*, *C*. The altitude corresponding to *C* is
, the median is , the bisector is . The radius of the
circumscribed circle is *R*, that of the inscribed circle is *r*.

Every triangle also has an **inscribed circle** tangent to its sides
and interior to the triangle (in other words, any three nonconcurrent
lines determine a circle). The center of this circle is the point of
intersection of the bisectors. We denote the radius of the inscribed
circle by *r*.

Every triangle has a **circumscribed circle** going through its
vertices; in other words, any three noncollinear points determine a
circle. The point of intersection of the medians is the center of
mass of the triangle (considered as an area in the plane). We denote
the radius of the circumscribed circle by *R*.

Introduce the following notations for an **arbitrary triangle** of
vertices *A*, *B*, *C* and sides *a*, *b*, *c* (see
Figure 1). Let , and be the lengths
of the altitude, bisector and median originating in vertex *C*, let
*r* and *R* be as usual the radii of the inscribed and circumscribed
circles, and let *s*=½(*a*+*b*+*c*). Then:

A triangle is **equilateral** if all its sides have the same length,
or, equivalently, if all its angles are the same (and equal to
60°). It is **isosceles** if two sides are the same, or,
equivalently, if two angles are the same. Otherwise it is
**scalene**.

For an **equilateral triangle** of side *a* we have:

area=¼*a*,

*r*=*a*,

*R*=*a*,

*h*=½*a*,

where *h* is any altitude. The altitude, the bisector and the median
for each vertex coincide.

For an **isosceles triangle**, the altitude for the unequal side is
also the corresponding bisector and median, but this is not true for
the other two altitudes. Many formulas for an isosceles triangle of
sides *a*, *a*, *c* can be immediately derived from those for a right
triangle of legs *a*, ½*c* (see Figure 2, left).

**Figure 2:** Left: An isosceles triangle can be divided into two congruent
right triangles. Right: notations for a right triangle.

For a **right triangle** the **hypothenuse** is the longest side
opposite the right angle; the **legs** are the two shorter sides,
adjacent to the right angle. The altitude for each leg equals the
other leg. In Figure 2 (right), *h* denotes the
altitude for the hypothenuse, while *m* and *n* denote the segments
into which this altitude divides the hypothenuse.

The following formulas apply for a right triangle:

*A*+*B*=90°

*r*=*ab*/(*a*+*b*+*c*)

*a*=*c* sin *A* = *c* cos *B*

*mc*=*b*

area=½*ab*

*c*=*a*+*b*
(Pythagoras)

*R*=½*c*

*b*=*c* sin *B* = *c* cos *A*

*nc*=*a*

*hc*=*ab*

The hypothenuse is a diameter of the circumscribed circle. The median
joining the midpoint of the hypothenuse (the center of the
circumscribed circle) to the right angle makes angles 2*A* and 2*B*
with the hypothenuse.

Additional facts about triangles:

- In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. This follows from the law of sines.
- (
**Ceva's Theorem**: see Figure 3, left.) In a triangle*ABC*, let*D*,*E*and*F*be points on the lines*BC*,*CA*and*AB*, respectively. Then the lines*AD*,*BE*and*CF*are concurrent if and only if the signed distances*BD*,*CE*, ... satisfyThis is so in three important particular cases: when the three lines are the medians, when they are the bisectors, and when they are the altitudes.

**Figure 3:**Left: Ceva's Theorem. Right: Menelaus's Theorem.

- (
**Menelaus's Theorem**: see Figure 3, right.) In a triangle*ABC*, let*D*,*E*and*F*be points on the lines*BC*,*CA*and*AB*, respectively. Then*D*,*E*and*F*are collinear if and only if the signed distances*BD*,*CE*, ... satisfy - Each side of a triangle is less than the sum of the other two. For any three lengths such that each is less than the sum of the other two, there is a triangle with these side lengths.

*Silvio Levy
Wed Oct 4 16:41:25 PDT 1995*

This document is excerpted from the 30th Edition of the *CRC Standard Mathematical Tables and Formulas* (CRC Press). Unauthorized duplication is forbidden.