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# 5.3: Medians and Altitudes in Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Define median and find the properties of the centroid.
• Apply medians to the coordinate plane.
• Construct the altitude of a triangle.

## Review Queue

1. Find the midpoint between (9, -1) and (1, 15).
2. Find the slope between (9, -1) and (1, 15). Then find the equation of the line.
3. Find the equation of the line that is perpendicular to the line from #2 through (-6, 2).

Know What? Triangles are frequently used in art. Your art teacher assigns an art project involving triangles. You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?

## Medians

Median: The line segment that joins a vertex and the midpoint of the opposite side (of a triangle).

LO¯¯¯¯¯¯¯\begin{align*}\overline{LO}\end{align*} is the median from L\begin{align*}L\end{align*} to the midpoint of NM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{NM}\end{align*}.

Example 1: Find the other two medians of LMN\begin{align*}\triangle LMN\end{align*}.

Solution: Find the midpoints of sides LN¯¯¯¯¯¯¯¯\begin{align*}\overline{LN}\end{align*} and LM¯¯¯¯¯¯¯¯¯\begin{align*}\overline{LM}\end{align*}, using a ruler. Be sure to always include the appropriate tick marks for the midpoints.

Centroid: The point of intersection for the medians of a triangle.

Investigation 5-5: Properties of the Centroid

Tools Needed: pencil, paper, ruler, compass

1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.

2. Draw in the medians and mark the centroid.

Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?

3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?

The properties discovered are summarized below.

Median Theorem: The medians of a triangle intersect at a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is the “balancing point” of a triangle.

If G\begin{align*}G\end{align*} is the centroid, then:

AGDGAnd:DG=23AD, CG=23CF, EG=23BE=13AD, FG=13CF, BG=13BE=12AG, FG=12CG, BG=12EG\begin{align*}AG &= \frac{2}{3} AD, \ CG = \frac{2}{3} CF, \ EG = \frac{2}{3} BE\\ DG &= \frac{1}{3} AD, \ FG = \frac{1}{3} CF, \ BG = \frac{1}{3} BE\\ \text{And}: \quad DG &= \frac{1}{2} AG, \ FG = \frac{1}{2} CG, \ BG = \frac{1}{2} EG\end{align*}

Example 2: I, K\begin{align*}I, \ K\end{align*}, and M\begin{align*}M\end{align*} are midpoints of the sides of HJL\begin{align*}\triangle HJL\end{align*}.

a) If JM=18\begin{align*}JM = 18\end{align*}, find JN\begin{align*}JN\end{align*} and NM\begin{align*}NM\end{align*}.

b) If HN=14\begin{align*}HN = 14\end{align*}, find NK\begin{align*}NK\end{align*} and HK\begin{align*}HK\end{align*}.

Solution:

a) JN=2318=12\begin{align*}JN = \frac{2}{3} \cdot 18 = 12\end{align*}. NM=JMJN=1812\begin{align*}NM = JM - JN = 18 - 12\end{align*}. NM=6.\begin{align*}NM = 6.\end{align*}

b) 14=23HK\begin{align*}14 = \frac{2}{3} \cdot HK\end{align*}

1432=HK=21\begin{align*}14 \cdot \frac{3}{2} = HK = 21\end{align*}. NK\begin{align*}NK\end{align*} is a third of 21, NK=7\begin{align*}NK = 7\end{align*}.

Example 3: Algebra Connection H\begin{align*}H\end{align*} is the centroid of ABC\begin{align*}\triangle ABC\end{align*} and DC=5y16\begin{align*}DC = 5y - 16\end{align*}. Find x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.

Solution:

12BH=HF3x+63x+68BH=2HF=2(2x1)=4x2=xHC=23DC32HC=DC 32(2y+8)=5y16 3y+12=5y16  28=2y\begin{align*}\frac{1}{2} BH= HF & \longrightarrow BH = 2HF && HC = \frac{2}{3} DC \longrightarrow \frac{3}{2} HC = DC\\ 3x + 6 &= 2(2x - 1) && \quad \ \frac{3}{2} (2y + 8) = 5y - 16\\ 3x + 6 &= 4x - 2 && \qquad \ 3y + 12 = 5y - 16\\ 8 &= x && \qquad \ \ \qquad 28 = 2y\end{align*}

## Altitudes

The last line segment within a triangle is an altitude. It is also called the height of a triangle.

Altitude: A line segment from a vertex and perpendicular to the opposite side. The red lines below are all altitudes.

When a triangle is a right triangle, the altitude, or height, is the leg. If the triangle is obtuse, then the altitude will be outside of the triangle.

Investigation 5-6: Constructing an Altitude for an Obtuse Triangle

Tools Needed: pencil, paper, compass, ruler

1. Draw an obtuse triangle. Label it ABC\begin{align*}\triangle ABC\end{align*}, like the picture to the right. Extend side AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, beyond point A\begin{align*}A\end{align*}.

2. Using Investigation 3-2, construct a perpendicular line to AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, through B\begin{align*}B\end{align*}.

The altitude does not have to extend past side AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, as it does in the picture. Technically the height is only the vertical distance from the highest vertex to the opposite side.

If you are constructing an altitude for an acute triangle, you may skip Step 1 of this investigation.

Know What? Revisited The point that you should put the wire through is the centroid. That way, each triangle will balance.

## Review Questions

• Questions 1-4 use Investigation 5-5.
• Questions 5-6 use Investigation 3-2 and 5-6.
• Questions 7-18 are similar to Examples 2 and 3.
• Questions 19-26 use review to discover something new.
• Questions 27-34 use the definitions of perpendicular bisector, angle bisector, median and altitude.
• Question 35 is similar to the proofs in the previous section.

Construction Construct the centroid for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-5.

1. Is the centroid always going to be inside of the triangle? Why?

Construction Construct the altitude from the top vertex for the following triangles. Trace each triangle onto a piece of paper and using Investigations 3-2 and 5-6.

For questions 7-11, B, D\begin{align*}B, \ D\end{align*}, and F\begin{align*}F\end{align*} are the midpoints of each side and G\begin{align*}G\end{align*} is the centroid. Find the following lengths.

1. If BG=5\begin{align*}BG = 5\end{align*}, find GE\begin{align*}GE\end{align*} and BE\begin{align*}BE\end{align*}
2. If CG=16\begin{align*}CG = 16\end{align*}, find GF\begin{align*}GF\end{align*} and CF\begin{align*}CF\end{align*}
3. If AD=30\begin{align*}AD = 30\end{align*}, find AG\begin{align*}AG\end{align*} and GD\begin{align*}GD\end{align*}
4. If GF=x\begin{align*}GF = x\end{align*}, find GC\begin{align*}GC\end{align*} and \begin{align*}CF\end{align*}
5. If \begin{align*}AG = 9x\end{align*} and \begin{align*}GD = 5x - 1\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}AD\end{align*}.

For questions 12-18, \begin{align*}N\end{align*} and \begin{align*}M\end{align*} are the midpoints of sides \begin{align*}\overline{XY}\end{align*}and \begin{align*}\overline{ZY}\end{align*}.

1. What is point \begin{align*}C\end{align*}?
2. If \begin{align*}XN = 5\end{align*}, find \begin{align*}XY\end{align*}.
3. If \begin{align*}XC = 6\end{align*}, find \begin{align*}XM\end{align*}.
4. If \begin{align*}ZN = 45\end{align*}, find \begin{align*}CN\end{align*}.
5. If \begin{align*}CM = 4\end{align*}, find \begin{align*}XC\end{align*}.
6. If \begin{align*}ZM = y\end{align*}, find \begin{align*}ZY\end{align*}.
7. If \begin{align*}ZN = 6x + 15\end{align*} and \begin{align*}ZC = 38\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}ZN\end{align*}.

Multistep Problem Find the equation of a median in the \begin{align*}x-y\end{align*} plane.

1. Plot \begin{align*}\triangle ABC: \ A(-6, 4), \ B(-2, 4)\end{align*} and \begin{align*}C(6, -4)\end{align*}
2. Find the midpoint of \begin{align*}\overline{AC}\end{align*}. Label it \begin{align*}D\end{align*}.
3. Find the slope of \begin{align*}\overline{BD}\end{align*}.
4. Find the equation of \begin{align*}\overline{BD}\end{align*}.
5. Plot \begin{align*}\triangle DEF: \ D(-1, 5), \ E(0, -1), \ F(6, 3)\end{align*}
6. Find the midpoint of \begin{align*}\overline{EF}\end{align*}. Label it \begin{align*}G\end{align*}.
7. Find the slope of \begin{align*}\overline{DG}\end{align*}.
8. Find the equation of \begin{align*}\overline{DG}\end{align*}.

Determine if the following statements are true or false.

1. The perpendicular bisector goes through the midpoint of a line segment.
2. The perpendicular bisector goes through a vertex.
3. The angle bisector passes through the midpoint.
4. The median bisects the side it intersects.
5. The angle bisectors intersect at one point.
6. The altitude of an obtuse triangle is inside a triangle.
7. The centroid is the balancing point of a triangle.
8. A median and a perpendicular bisector intersect at the midpoint of the side they intersect.

Fill in the blanks in the proof below.

1. Given: Isoscles \begin{align*}\triangle ABC\end{align*} with legs \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{AC}\end{align*} \begin{align*}\overline{BD} \perp \overline{DC}\end{align*} and \begin{align*}\overline{CE} \perp \overline{BE}\end{align*} Prove: \begin{align*}\overline{BD} \cong \overline{CE}\end{align*}
Statement Reason

1. Isosceles \begin{align*}\triangle ABC\end{align*} with legs \begin{align*} \overline{AB}\end{align*} and \begin{align*}\overline{AC}\end{align*}

\begin{align*}\overline{BD} \perp \overline{DC} \end{align*} and \begin{align*}\overline{CE} \perp \overline{BE}\end{align*}

2. \begin{align*}\angle DBC \cong \angle ECB\end{align*}
3. Definition of perpendicular lines
4. \begin{align*}\angle BEC \cong \angle CEB\end{align*}
5. Reflexive PoC
6. \begin{align*}\triangle BEC \cong \triangle CDB\end{align*}
7. \begin{align*}\overline{BD} \cong \overline{CE}\end{align*}

1. \begin{align*}midpoint=\left (\frac{9+1}{2},\frac{-1+15}{2}\right ) = (5,7)\end{align*}
2. \begin{align*}m=\frac{15+1}{1-9}=\frac{16}{-8}=-2 \qquad \qquad 15=-2(1)+b \qquad \qquad y=-2x+17\!\\ {\;}\qquad \qquad \qquad \qquad \qquad \qquad \ \ 17=b\end{align*}
3. \begin{align*}y=\frac{1}{2} x+b\!\\ 2= \frac{1}{2} (-6) + b\!\\ 2=-3+b\!\\ 5=b\!\\ y=\frac{1}{2} x+5\end{align*}

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