The altitudes, medians and angle bisectors of a triangle are important special lines with unique properties. This page presents their definitions and problems with solutions.
An altitude of a triangle is a line segment through a vertex that is perpendicular to the opposite side. The three altitudes of any triangle intersect at a single point called the orthocenter, which may be inside or outside the triangle.
Example of triangle with orthocenter inside:
Example of triangle with orthocenter outside:
A median of a triangle is a line segment from a vertex to the midpoint of the opposite side. The three medians intersect at the centroid, which is the triangle's center of mass.
For a triangle with vertices \(A(x_A, y_A)\), \(B(x_B, y_B)\), and \(C(x_C, y_C)\), the centroid coordinates are:
\[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) \]An angle bisector divides an interior angle into two equal angles. The three angle bisectors intersect at the incenter, which is the center of the inscribed circle.
For a triangle with sides of lengths \(a\), \(b\), \(c\) opposite vertices \(A(x_A, y_A)\), \(B(x_B, y_B)\), and \(C(x_C, y_C)\) respectively, the incenter coordinates are:
\[ I = \left( \frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c}, \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c} \right) \]Problem 1
Find the centroid of triangle with vertices \(A(-2, 0)\), \(B(4, 3)\), and \(C(1, 6)\).
Problem 2
Find the incenter of triangle with vertices \(A(-1, 0)\), \(B(3, 3)\), and \(C(3, -3)\).
Problem 3
Find the orthocenter of triangle with vertices \(A(0, 0)\), \(B(5, 0)\), and \(C(3, 3)\).
Solution to Problem 1
Using the centroid formula:
\[
x_G = \frac{-2 + 4 + 1}{3} = 1, \quad y_G = \frac{0 + 3 + 6}{3} = 3
\]
Thus the centroid is at \(G(1, 3)\).
Solution to Problem 2
First calculate side lengths:
\(a = BC = \sqrt{(3-3)^2 + (3-(-3))^2} = 6\)
\(b = CA = \sqrt{(-1-3)^2 + (0-(-3))^2} = 5\)
\(c = AB = \sqrt{(-1-3)^2 + (0-3)^2} = 5\)
Using the incenter formula:
\[
x_I = \frac{6(-1) + 5(3) + 5(3)}{6+5+5} = \frac{24}{16} = \frac{3}{2}
\]
\[
y_I = \frac{6(0) + 5(3) + 5(-3)}{16} = 0
\]
Thus the incenter is at \(I\left(\frac{3}{2}, 0\right)\).
Solution to Problem 3
The orthocenter is the intersection of the altitudes.
1. Altitude through \(C\): Since \(AB\) is horizontal (slope 0), the altitude through \(C(3,3)\) is vertical: \(x = 3\).
2. Altitude through \(A\): Slope of \(BC = \frac{3-0}{3-5} = -\frac{3}{2}\).
Perpendicular slope = \(\frac{2}{3}\).
Equation through \(A(0,0)\): \(y = \frac{2}{3}x\).
Intersection of \(x=3\) and \(y=\frac{2}{3}x\): \(y = \frac{2}{3}(3) = 2\).
Thus the orthocenter is at \(H(3, 2)\).
