What if you were given one or more of a triangle's angle measures? How would you determine where the triangle's altitude would be found? After completing this Concept, you'll be able to answer this type of question.

### Watch This

CK-12 Foundation: Chapter5AltitudesA

James Sousa: Altitudes of a Triangle

### Guidance

An **altitude** is a line segment in a triangle from a vertex and perpendicular to the opposite side, it is also known as the height of a triangle. All of the red lines are examples of altitudes:

As you can see, an altitude can be a side of a triangle or outside of the triangle. When a triangle is a right triangle, the altitude, or height, is the leg. To construct an altitude, construct a perpendicular line through a point not on the given line. Think of the vertex as the point and the given line as the opposite side.

##### Investigation: Constructing an Altitude for an Obtuse Triangle

Tools Needed: pencil, paper, compass, ruler

- Draw an obtuse triangle. Label it \begin{align*}\triangle ABC\end{align*}, like the picture to the right. Extend side \begin{align*}\overline{AC}\end{align*}, beyond point \begin{align*}A\end{align*}.
- Construct a perpendicular line to \begin{align*}\overline{AC}\end{align*}, through \begin{align*}B\end{align*}.

The altitude does not have to extend past side \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.

As was true with perpendicular bisectors, angle bisectors, and medians,the altitudes of a triangle are also concurrent. Unlike the other three, the point does not have any special properties.

**Orthocenter:** The point of concurrency for the altitudes of triangle.

Here is what the orthocenter looks like for the three triangles. It has three different locations, much like the perpendicular bisectors.

Acute Triangle |
Right Triangle |
Obtuse Triangle |
---|---|---|

The orthocenter is inside the triangle. | The legs of the triangle are two of the altitudes. The orthocenter is the vertex of the right angle. | The orthocenter is outside the triangle. |

#### Example A

Which line segment is an altitude of \begin{align*}\triangle ABC\end{align*}?

In a right triangle, the altitude, or the height, is the leg. If we rotate the triangle so that the right angle is in the lower left corner, we see that leg \begin{align*}BC\end{align*} is an altitude.

#### Example B

A triangle has angles that measure \begin{align*}55^\circ, 60^\circ,\end{align*} and \begin{align*}65^\circ\end{align*}. Where will the orthocenter be found?

Because all of the angle measures are less than \begin{align*}90^\circ\end{align*}, the triangle is an acute triangle. The orthocenter of any acute triangle is inside the triangle.

#### Example C

A triangle has an angle that measures \begin{align*}95^\circ\end{align*}. Where will the orthocenter be found?

Because \begin{align*}95^\circ > 90^\circ\end{align*}, the triangle is an obtuse triangle. The orthocenter will be outside the triangle.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter5AltitudesB

### Guided Practice

1. True or false: The altitudes of an obtuse triangle are inside the triangle.

2. Draw the altitude for the triangle shown.

3. Draw the altitude for the triangle shown.

**Answers:**

1. Every triangle has three altitudes. For an obtuse triangle, at least one of the altitudes will be outside of the triangle, as shown in the picture at the beginning of this concept.

2. The triangle is an acute triangle, so the altitude is inside the triangle as shown below so that it is perpendicular to the base.

3. The triangle is a right triangle, so the altitude is already drawn. The altitude is \begin{align*}\overline{XZ}\end{align*}.

### Interactive Practice

### Explore More

Write a two-column proof.

- Given: 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} \bot \overline{DC}\end{align*} and \begin{align*}\overline{CE} \bot \overline{BE}\end{align*} Prove: \begin{align*}\overline{BD} \cong \overline{CE}\end{align*}

For the following triangles, will the altitudes be inside the triangle, outside the triangle, or at the leg of the triangle?

- \begin{align*}\triangle JKL\end{align*} is an equiangular triangle.
- \begin{align*}\triangle MNO\end{align*} is a triangle in which two the angles measure \begin{align*}30^\circ\end{align*} and \begin{align*}60^\circ\end{align*}.
- \begin{align*}\triangle PQR\end{align*} is an isosceles triangle in which two of the angles measure \begin{align*}25^\circ\end{align*}.
- \begin{align*}\triangle STU\end{align*} is an isosceles triangle in which two angles measures \begin{align*}45^\circ\end{align*}.

Given the following triangles, which line segment is the altitude?