Altitudes and Orthocenter of a Triangle

The altitude of a triangle is the line that passes through a vertex and is perpendicular to the side (or to the extension of that side) opposite to that vertex.

The problem presented here is about the point of concurrency of the altitudes of a triangle. This point is referred to as the orthocenter of the triangle.

For a right triangle, this point of concurrency of the altitudes is at one of its vertices. This is so because two of its altitudes coincide with its legs. See illustration below.

For an obtuse triangle, this point of concurrency of the altitudes is outside of the triangle. Two of its altitudes are drawn to the respective extensions of its opposite sides. See illustration below.

The Problem

Given triangle ABC with its vertices at A(-5, 0), B(3, 4) and C(0, -5), find the point of concurrency of its altitudes.

Discussion

The altitudes of a triangle are concurrent at a point.This point of concurrency is called the orthocenter of a triangle.

In the illustrated triangle above, triangle ABC is a scalene triangle. Line segments AT, BU and CS are its altitudes. Point O is its orthocenter.

Solution

Point of Concurrency: O (-2, -1)

Distance Between Two Points

• distance_AB = √ 80
• distance_BC = √ 90
• distance_AC = √ 50
• distance_AT = √ 40
• distance_BU = √ 72
• distance_CS = √ 45

Equations of the Line Segments

• side AB: x - 2y + 5 = 0; x-intercept = - 5; y-intercept = 5/2
• side BC: 3x - y - 5 = 0; x-intercept = 5/3; y-intercept = - 5
• side AC: x + y + 5 = 0; x-intercept = - 5; y-intercept = - 5
• altitude AT: x + 3y + 5 = 0; x-intercept = - 5; y-intercept = - 5/3
• altitude BU: x - y + 1 = 0; x-intercept = -1; y-intercept = 1
• altitude CS: 2x + y + 5 = 0; x-intercept = - 5/2; y-intercept = - 5

Slopes of the Line Segments

• m_AB = 1/2
• m_BC = 3
• m_AC = - 1
• m_AT = - 1/3
• m_BU = 1
• m_CS = - 2

Medians And Centroid Of A Triangle

This is a continuation of my previous post. The problems presented here are about the medians of a triangle and their point of concurrency.

The following theorems are applied in the given problems below:

• In a right triangle,the length of the median to the hypotenuse is one half the length of the hypotenuse (see Problem 1).
• In an isosceles triangle, the median to the base is perpendicular to the base itself (see Problem 2).
• The medians of a triangle have a point of concurrency (see Problem 2).

Problem 1

Given a right triangle ABC with vertices at A(-2, 5), B(2, -5) and C(-5, 2) and its median that intersects the hypotenuse, find the length of hypotenuse AB if median CY = √29.

Discussion

Point Y is the midpoint of hypotenuse AB.

Line segment CY is a median, a line drawn from a vertex to the midpoint of the opposite side.

Solution

Distance Between Two Points

• distance_AC = √ 18
• distance_AB = √ 116
• distance_BC = √ 98
• distance_CY = distance_AY = distance_BY = √ 29

Equations of the Line Segments

• line segment AB: 5x + 2y = 0; x-intercept = 0; y-intercept = 0
• line segment BC: x + y + 3 = 0; x-intercept = -3; y-intercept = -3
• line segment AC: x - y + 7 = 0; x-intercept = -7; y-intercept = 7
• line segment CY: 2x + 5y = 0; x-intercept = 0; y-intercept = 0

Slopes of the Line Segments

• m_AB = - 5/2
• m_BC = - 1
• m_AC = 1
• m_CY = - 2/5

Problem 2

Given isosceles triangle ABC with its medians AE, BF and CD, find their point of concurrency. The vertices of the triangle are A(-6, 5), B(4, 1) and C(-2, -5).

Discussion

In the above figure, the medians of isosceles triangle ABC intersect at point X which is called centroid and is the center of gravity of the triangle.

This point is the trisection point of each median.

Taking median AE, line segment AX = 2/3 AE = 2 EX and line segment EX = 1/3 AE. This property holds true for the other two medians as well.

Solution

Point of Concurrency of the Medians: X (- 4/3, 1/3)

Distance Between Two Points

• distance_AB = distance_AC = √ 116
• distance_BC = √ 72
• distance_AE = √ 98
• distance_BF = distance_CD = √ 65
• distance_AX = 2/3 distance = √ 392/9
• distance_EX = 1/3 distance_AE = √ 98/9
• distance_BX = 2/3 distance_BF = √ 260/9
• distance_FX = 1/3 distance_BF = √ 65/9
• distance_CX = 2/3 distance_CD = √ 260/9
• distance_DX = 1/3 distance_CD = √ 65/9

Midpoints of the Sides

• midpoint of side AB: D(-1, 3)
• midpoint of side BC: E(1, -2)
• midpoint of side AC: F(-4, 0)

Equations of the Line Segments

• line segment AB: 2x + 5y - 13 = 0; x-intercept = 13/2; y-intercept = 13/5
• line segment BC: x - y - 3 = 0; x-intercept = 3; y-intercept = - 3
• line segment AC: 5x + 2y + 20 = 0; x-intercept = - 4; y-intercept = - 10
• line segment AE: x + y + 1 = 0; x-intercept = - 1; y-intercept = - 1
• line segment BF: x - 8y + 4 = 0; x-intercept = - 4; y-intercept = 1/2
• line segment CD: 8x - y + 11 = 0; x-intercept = - 11/8; y-intercept = 11

Slopes of the Line Segments

• m_AB = - 2/5
• m_BC = 1
• m_AC = - 5/2
• m_AE = -1

Parts Of A Triangle

Aside from its vertices, sides (legs and base), interior and exterior angles, a triangle has also the following parts:

• median
• altitude
• perpendicular bisector of its sides
• angle bisector

These parts are discussed in the following problem.

The Problem

Given isosceles triangle ABC with vertices at A(-3, 4), B(5, 0) and C(1, -4), show analytically that the perpendicular bisector of its base passes through its vertex angle.

Discussion

The illustration above shows isosceles triangle ABC with point X as the midpoint of side BC which is the base.

Given the definitions of the following parts:

• The median is the line formed by a triangle's vertex and the midpoint of the side opposite to that vertex.
• The altitude is the perpendicular line drawn from a triangle's vertex to the side opposite to that vertex (or to the extension of the side opposite to that vertex).
• The perpendicular bisector of a triangle's side divides that side into two equal line segments.
• The angle bisector is a line that divides an angle into two equal angles.

line segment AX is all of the above.

The solution to the above problem serves as a coordinate proof of the theorem: the perpendicular bisector of the base of an isosceles triangle divides the vertex angle into two equal angles.

Solution

In order to show that the perpendicular bisector of the base passes through the vertex angle, it must be shown that line segment AX:

• is perpendicular to the base AB;
• divides base AB into two equal line segments;
• divides angle A into two equal angles.

Distance Between Two Points

• distance_AB = distance_AC = √ 80
• distance_BC = √ 32
• distance_BX = distance_CX = √ 8

Equations of the Line Segments

• line segment AB: x + 2y - 5 = 0; x-intercept = 5; y-intercept = 5/2
• line segment BC: x - y - 5 = 0; x-intercept = 5; y-intercept = - 5
• line segment AC: 2x + y + 2 = 0; x-intercept = - 1; y-intercept = - 2
• line segment AX: x + y - 1 = 0; x-intercept = 1; y-intercept = 1

Slopes of the Line Segments

• m_AB = - 1/2
• m_BC = 1
• m_AC = - 2
• m_AX = - 1

Interior Angles of the Triangle

• vertex angle A = 36.87°
• base angle B = base angle C = 71.56°
• angle CAX = angle BAX = 18.43°

Medians And Altitudes of Isosceles Triangles

Given the same isosceles triangle ABC:

• point D is the midpoint of leg AB;
• point E is the midpoint of leg AC;
• point F is the intersection of leg AB and altitude CF;
• point G is the intersection of leg AC and altitude BG.

Line segments CD and BE are medians to the legs of triangle ABC.



The graph above shows that for any isosceles triangle, its medians to the legs are congruent.

Distance Between Two Points

• distance_AD = distance_BE = distance_CD = distance_CE = √ 20
• distance_BE = distance_CD = √ 36
• distance_BG = distance_CF = √ 720/25

Equations of the Medians And the Altitudes

• median CD: x = 1
• median BE: y = 0
• altitude CF: 2x - y - 6 = 0
• altitude BG: x - 2y - 5 = 0

Coordinates of:

• point F: (17/5, 4/5)
• point G: (1/5, - 12/5)

This last graph illustrates that for any isosceles triangle, its altitudes to the legs are congruent.