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Wize High School Geometry Textbook (Common Core) > Triangles

Properties of Triangles

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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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Construct the centroid of a triangle


Our goal in this construction is to draw the centroid of a given triangle. The centroid is the intersection of all three medians of a triangle.

Using Technology


Begin with triangle △ABC\triangle ABC
  1. Construct the midpoint for side AB‾\overline{AB}. Label this point as point DD.
  2. Draw the line segment DC‾\overline{DC}.
  3. Repeat steps 1 and 2 for the other two sides. Label the intersection of all of these segments as point GG.
From this we now have that GG is the centroid of triangle △ABC\triangle ABC.
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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.
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Example: Properties of Triangles

Based on the following angle measurements, come up with a conjecture for the relationship between exterior and interior angles of any triangle. Then try to disprove the conjecture by finding a counterexample, or actually prove the conjecture using angle properties of triangles.

Right angle triangle


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Isosceles Triangle



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Scalene Triangle


According to the 3 triangles given, it seems like the exterior angle is the sum of the two opposite interior angles.
Although we have 3 examples here that support our conjecture, it's not enought to actually prove the conjecture because "what if there is an example out there that disproves this conjecture?"

But we can really prove this conjecture by not looking at examples at all. Instead, we turn to properties and facts about angles that we already know to be true.

Let's take a look at any triangle, we can label 4 angles as a, b, c\bct{a,\ b,\ c} and d\bct d:

Recall:
  1. Supplemental angles that form a straight line must add up to 180°180\degree
  2. The sum of the interior angles in any triangle is 180°180\degree
Using property 1: c+d=180°\colorbox{yellow}{$c+d=180\degree$}
Using property 2: a+b+d=180°\colorbox{yellow}{$a+b+d=180\degree$}

Since c+dc+d and a+b+da+b+d both equal 180°180\degree, we know that c+dc+d and a+b+da+b+d must equal each other:
c+d=a+b+d−d=            −dc=a+b\begin{array}{rcl} c+d&=&a+b+d\\ \scriptsize-d&=&~~~~~~~~~~~~\scriptsize-d\\ c&=&a+b \end{array}
We see that for any triangle, c=a+b\boxed{c=a+b}, meaning that the exterior angle of any triangle is the sum of the two opposite interior angles!