Find a piece of cardstock or thick paper. Use a ruler and pencil to draw a fairly large random triangle on the paper. Use your ruler to help you to construct the centroid of the triangle. Carefully cut out the triangle and try to balance it on the tip of your pencil. Where is the balancing point?

#### Watch This

http://www.youtube.com/watch?v=aaIX1rUdrgs James Sousa: Using the Properties of the Medians of a Triangle to Solve for Unknown Values

#### Guidance

There are nine theorems related to triangles that are helpful to know.

- The sum of the measures of the interior angles of a triangle is \begin{align*}180^\circ\end{align*}.
- The base angles of an isosceles triangle are congruent.
- If a triangle has two congruent angles then it is isosceles. (Note that this is the converse of #2)
- The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles.
- The segment connecting two midpoints of a triangle is both parallel to and one half the length of the third side of the triangle.
- The three medians of a triangle meet at a point called the centroid. The centroid divides each median in a 2:1 ratio with the larger segment being the one from the vertex to the centroid.
- The three angle bisectors of a triangle meet at a point called the incenter.
- The three altitudes of a triangle meet at a point called the orthocenter.
- The three perpendicular bisectors of a triangle meet at a point called the circumcenter.

**Example A**

Point \begin{align*}P\end{align*} is the centroid of \begin{align*}\Delta ABC\end{align*}. Find the length of \begin{align*}\overline{AP}\end{align*}.

**Solution:** Because \begin{align*}P\end{align*} is the centroid, it divides each median in a 2:1 ratio, where the length of the segment from the vertex to the centroid is the longer segment. This means that \begin{align*}\overline{AP}\end{align*} is twice the length of the segment that is marked as \begin{align*}2 \ in\end{align*}. Therefore, \begin{align*}AP=4\end{align*}.

**Example B**

Solve for \begin{align*}x\end{align*}.

**Solution:** The sum of the measures of the angles is \begin{align*}180^\circ\end{align*}. Set up an equation and solve for \begin{align*}x\end{align*}.

\begin{align*}3x+15+3x-5+2x+2 &=180\\ 8x+12 &=180\\ 8x &=168\\ x &=21 \end{align*}

**Example C**

Prove that two of the medians of an isosceles triangle are congruent.

**Solution:** You can complete this proof using the triangle below. Your goal is to prove that \begin{align*}\Delta DBC\cong \Delta ECB\end{align*} and then show that the medians are congruent because they are corresponding parts of the triangles.

Consider isosceles \begin{align*}\Delta ABC\end{align*} with \begin{align*}\overline{AB}\cong \overline{AC}\end{align*}. Midpoints \begin{align*}D\end{align*} and \begin{align*}E\end{align*} divide \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{AC}\end{align*} respectively, so \begin{align*}\overline{AD}\cong\overline{DB}\cong\overline{AE}\cong\overline{EC}\end{align*}. Because it is an isosceles triangle, \begin{align*}\angle ABC\cong \angle ACB\end{align*}. \begin{align*}\overline{BC}\cong \overline{BC}\end{align*} because any segment is congruent to itself. Therefore, \begin{align*}\Delta DBC\cong \Delta ECB\end{align*} by \begin{align*}SAS\cong\end{align*}. \begin{align*}\overline{DC}\cong \overline{EB}\end{align*} because corresponding parts of congruent triangles are congruent.

**Concept Problem Revisited**

You should find that the centroid is the balancing point of the triangle. This means that the centroid is the center of gravity for the triangle when constructed in real life.

#### Vocabulary

The ** median** of a triangle is a line segment that connects the

**of one side of the triangle with the opposite**

*midpoint***. The three medians of a triangle meet at a point called the**

*vertex*

*centroid.*
The ** altitude** of a triangle is a line that is

**to one side of the triangle and passes through the opposite**

*perpendicular***. The three altitudes of a triangle meet at a point called the**

*vertex***.**

*orthocenter*
A ** perpendicular bisector** of a segment is a line that bisects the segment and meets the segment at a right angle. The three perpendicular bisectors of a triangle meet at a point called the

**.**

*circumcenter*
An ** angle bisector** of an angle is a line that bisects the angle. The three angle bisectors of a triangle meet at a point called the

**.**

*incenter*
A ** point of concurrency** is the intersection point of three or more lines.

A ** scalene triangle** has no sides that are congruent.

An ** isosceles triangle** has two sides that are congruent.

An ** equilateral triangle** has three sides that are congruent.

A ** right triangle** has one right angle.

An ** obtuse triangle** has one angle with a measure that is greater than \begin{align*}90^\circ\end{align*}.

An ** acute triangle** has all three angles with measures less than \begin{align*}90^\circ\end{align*}.

An ** equiangular triangle** has three congruent angles.

#### Guided Practice

1. Point \begin{align*}P\end{align*} is the centroid of \begin{align*}\Delta ABC\end{align*} and \begin{align*}BD=18\end{align*}. Find the length of \begin{align*}\overline{BP}\end{align*}.

2. The three altitudes of a triangle meet at a point called the ____________.

3. Solve for \begin{align*}x\end{align*}.

**Answers:**

1. The centroid divides the median in a 2:1 ratio where \begin{align*}\overline{BP}\end{align*} is the longer segment.

\begin{align*}2x+1x &=18 \\ x &=6\\ 2x &=12\end{align*}

The two segments have lengths of 6 and 12, so \begin{align*}BP=12\end{align*}.

2. orthocenter

3. The measure of the exterior angle is equal to the sum of the measures of the remote interior angles. Set up an equation to solve for \begin{align*}x\end{align*}.

\begin{align*}11x-15 &=4x-2+6x+2\\ 11x-15 &=10x\\ x &=15 \end{align*}

#### Practice

1. Point \begin{align*}P\end{align*} is the centroid of \begin{align*}\Delta ABC\end{align*} and \begin{align*}CD=12\end{align*}. Find the lengths of \begin{align*}\overline{CP}\end{align*} and \begin{align*}\overline{DP}\end{align*}.

2. What are the four points of concurrency for a triangle?

3. The three medians of a triangle meet at a point called the ____________.

4. The three angle bisectors of a triangle meet at a point called the _____________.

5. The three perpendicular bisectors of a triangle meet at a point called the _____________.

6. Investigate which points of concurrency are always **inside** a triangle and which points of concurrency are sometimes **outside** a triangle (use geometry software to help). What did you find out?

7. Explore the points of concurrency for an equilateral triangle (use geometry software to help). What do you notice?

Solve for \begin{align*}x\end{align*}.

8.

9.

10.

11. Solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

12. Solve for \begin{align*}x\end{align*}.

In the triangle below, \begin{align*}BC=x\end{align*}, \begin{align*}AE=x-2\end{align*}, \begin{align*}D\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*}, and \begin{align*}E\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}, and the perimeter of the triangle is 42.

13. Solve for \begin{align*}x\end{align*}.

14. Find \begin{align*}DE\end{align*}, \begin{align*}BC\end{align*}, \begin{align*}AE\end{align*}, and \begin{align*}AD\end{align*}.

15. A kite has diagonals with lengths 4 and 6. An inner quadrilateral is formed by joining the midpoints of each of the four sides of the kite. What is the perimeter of this inner quadrilateral?