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Wize High School Grade 10 Math Textbook > Analytic Geometry: Geometric Properties

Properties of Triangles

Classifying TrianglesPrevious Section
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Line Segments and Centers in a Triangle

Median

  • Median: a line that connects a vertex to the midpoint of its opposite side
  • Centroid: where the 3 medians in a triangle meet
  • *Each median bisects the area of the triangle


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Altitude

  • Altitude: a line that connects a vertex to its opposite side, and is perpendicular to the opposite side
  • An altitude is often called the "height" of the triangle
  • Orthocenter: where the 3 altitudes in a triangle meet
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Angle Bisector

  • Angle bisector: a line that connects a vertex to its opposite side, and it cuts the angle in half
  • Incenter: where the 3 angle bisectors in a triangle meet


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Right Bisector

  • Perpendicular (right) bisector: a line that cuts a side in the triangle in half, and is perpendicular to that side
  • Circumcenter: where the 3 right bisectors in a triangle meet
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Wize Tip
In an equilateral triangle, the median, altitude, and right bisector from a single vertex are all the same!

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Example: Median of a Triangle

A triangle has vertices A(1,3)A(1,3), B(3,5)B(3,5), and C(5,1)C(5,1).

a) Find the length of the median that connects vertex C to its opposite side.

Recall: a median connects a vertex to the midpoint on the opposite side.

Midpoint of AB:
MAB=(x1+x22,y1+y22)          =(1+32,3+52)          =(2,4)\begin{array}{ccccc} M_{AB}=\Large(&\dfrac{x_1+x_2}{2}&,&\dfrac{y_1+y_2}{2}&\Large)\\[1em] ~~~~~~~~~~=\Large(&\dfrac{1+3}{2}&,&\dfrac{3+5}{2}&\Large)\\[1em] ~~~~~~~~~~=\Large(&2&,&4&\Large)\\[1em] \end{array}

Length of Median (length between MABM_{AB} and vertex C:
(x2−x1)2+(y2−y1)2=(5−2)2+(1−4)2=(3)2+(−3)2=9+9=18\begin{array}{rcl} &&\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\[1em] &=&\sqrt{(5-2)^2+(1-4)^2}\\[1em] &=&\sqrt{(3)^2+(-3)^2}\\[1em] &=&\sqrt{9+9}\\[1em] &=&\sqrt{18} \end{array}
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b) Show that the median you found in a) is also an altitude.

The median is an altitude if it's perpendicular to the line segment AB.

mAB=y2−y1x2−x1=5−33−1=22=1mMedian=y2−y1x2−x1=1−45−2=−33=−1\begin{array}{c|c} \begin{array}{rcl} m_{AB}&=&\dfrac{y_2-y_1}{x_2-x_1}\\[1em] &=&\dfrac{5-3}{3-1}\\[1em] &=&\dfrac{2}{2}\\[1em] &=&1 \end{array} & \begin{array}{rcl} m_{Median}&=&\dfrac{y_2-y_1}{x_2-x_1}\\[1em] &=&\dfrac{1-4}{5-2}\\[1em] &=&\dfrac{-3}{3}\\[1em] &=&-1 \end{array} \end{array}

Since these slopes are negative reciprocal, we know that the median is perpendicular to the side AB. So, the median is also the altitude.

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c) Determine the area of triangle △ABC\triangle ABC.

Let's use AB as the base of the triangle. Since the median from the vertex C is also the altitude, the median is the height.
  • Length of AB: (3−1)2+(5−3)2=8\sqrt{(3-1)^2+(5-3)^2}=\sqrt {8}
  • Length of median: 18\sqrt{18}

Area=base×height2=Length of AB×Length of median2=8182=1442=122=6\begin{array}{rcl} Area&=&\dfrac{base\times height}{2}\\[1em] &=&\dfrac{\text{Length of AB}\times\text{Length of median}}{2}\\[1em] &=&\dfrac{\sqrt{8}\sqrt{18}}{2}\\[1em] &=&\dfrac{\sqrt{144}}{2}\\[1em] &=&\dfrac{12}{2}\\[1em] &=&6 \end{array}

Practice: Properties of Triangles

A triangle has vertices A(2,4)A\left(2,4\right), B(6,6)B\left(6,6\right), and C(8,2)C\left(8,2\right).

a) Determine the centroid of the triangle.

b) Show that this centroid divides the medians in a 2:1 ratio

Practice: Properties of Triangles

A triangle has vertices A(2,4)A\left(2,4\right), B(6,6)B\left(6,6\right), and C(8,2)C\left(8,2\right).
a) Calculate the slopes of each side of the triangle.

b) A midsegment is a line segment that connects two midpoints. Determine the slopes of the midsegments between 2 sides of the triangle. (*there should be 3 midsegments, each joining the midpoints between 2 sides of the triangle)

c) Make an observation about the slopes of the sides and the slopes of the midsegments.