Altitudes and the Orthocenter

Overview

Altitudes are perpendicular segments drawn from a vertex to the opposite side (or its extension). They play a central role in triangle geometry because they connect directly to area and often create right triangles, similar triangles, cyclic quadrilaterals, and useful length relationships.

The three altitudes of a triangle intersect at a single point called the orthocenter. The location of the orthocenter depends entirely on the type of triangle.

Definitions and Key Ideas

An altitude of a triangle is a line through a vertex perpendicular to the opposite side (or its extension).

It can be proven that the three altitudes of a triangle intersect at one point; this is called the orthocenter.

• In a triangle, area $K = \frac{1}{2}ah_a$ (etc.), so $h_a=\frac{2K}{a}, h_b=\frac{2K}{b}, h_c=\frac{2K}{c}.$

• In a triangle, $\frac{h_a}{h_b}=\frac{b}{a},$ and similarly for any pair of altitudes.

• Multiple equal angles and similar triangles are produced by the altitudes, found via angle chasing and cyclic quadrilaterals.

• Quadrilaterals such as $A H_C H H_B$ and $B H_C H_B C$ are cyclic.

• In a right triangle, the following properties hold:

• In an acute triangle, the orthocenter is inside the triangle.
• In a right triangle, the orthocenter is the right-angle vertex.
• In an obtuse triangle, the orthocenter lies outside the triangle.

Properties

Let $\triangle ABC$ have side lengths $a, b, c$ opposite vertices $A,B,C,$ with corresponding altitudes $h_a, h_b, h_c.$ Let the feet of the altitudes be $H_A, H_B, H_C.$

Relationship to Area

Because any side may be chosen as a base for a triangle, the area $K$ satisfies

Corollary 1: $h_a=\frac{2K}{a}, h_b=\frac{2K}{b}, h_c=\frac{2K}{c}$

Corollary 2: $ah_a=bh_b=ch_c=2K,$ and so

and similarly for any pair of altitudes.

Equal Angles and Similar Triangles

Dropping altitudes creates equal $90^\circ$ angles, giving rise to numerous similar triangles by AA similarity. Below is a visualization of many pairs of equal angles:

For instance, $\triangle AH_AC \sim \triangle BH_BC,$ which is an alternative way to prove Corollary 2. Let $H$ be the orthocenter of $ABC.$ Then

and similarly for other pairs of vertices, such as $A, C$ and $B,C.$

Right Triangle Altitudes

Dropping an altitude often creates similar triangles, and one especially important case occurs in a right triangle. Suppose $ABC$ is a right triangle with right angle at $C$, and altitude $CD$ is drawn from $C$ to hypotenuse $AB$.

Then

This leads to several useful relationships:

These formulas appear regularly in AMC geometry problems.

Connection: These relationships can also be proven using Power of a Point. Considering the first equation, reflect $C$ across hypotenuse $AB$ to point $C',$ to yield a cyclic quadrilateral $AC' B C.$ Then $(CD)(C'D) = (AD)(DB),$ and the proof follows from how $C'D=CD.$ Similarly, reflect $D$ across $AC$ to $D'$ and consider cyclic quadrilateral $AD'CD.$ It will have diameter $AC,$ so we know $BC$ is a tangent to its circumcircle, since $BC \perp AC.$ Then use the tangent/secant statement of PoAP to obtain $BC^2 = BD \cdot BA.$ The proof of the second statement is similar.

Angle Chasing and Cyclic Quadrilaterals

We'll use $\angle A$ to refer to the largest angle $\angle BAC,$ and so on. It holds that

and likewise for the other two vertices.

Proof: Use $180 - \angle H_B H C$ or $360 - (\angle A + \angle A H_C C + \angle A H_B H).$ Or, use the fact that $A H_C H H_B$ is cyclic!

Notice that $A H_C H H_B$ is cyclic, and so is $CH_B H H_A$ and $B H_A H H_C,$ since opposite angles sum to $180^\circ.$ Let's only angle chase on $A H_C H H_B$ and generalize by symmetry. Angles that subtend the same arc are equal, so $\angle H_C H_B H = \angle H_C A H = 90 - \angle B,$ and $\angle H H_C H_B = \angle HA H_B = 90 - \angle C.$ Similarly, $\angle A H H_B = \angle A H_C H_B = \angle C,$ and $\angle A H H_C = \angle A H_B H_C = \angle B.$

Note: The quadrilaterals $BH_C H_B C, A H_C H_A C, A H_B H_A B$ are also cyclic, since opposite angles formed by a diagonal and side are equal. Circumcircles of two cyclic quadrilaterals have been drawn in the below diagram to help visualize this.

If we continue to chase equal angles, we obtain the following diagram, where red angles $= 90 - \angle C,$ blue angles $= 90 - \angle A,$ and green angles $= 90 - \angle B.$

Configurations involving the feet of altitudes and the orthocenter frequently lead to cyclic quadrilaterals and angle-chasing opportunities. Whenever you see multiple altitudes, look for right angles and consider whether four points lie on a common circle.

Triangle $\triangle H_A H_B H_C$ is called the orthic triangle of $\triangle ABC,$ and the diagram reveals a really nice property: the altitudes of $\triangle ABC$ are the angle bisectors of the orthic triangle! Also, the interior angles of the orthic triangle are

which indeed sum to $540 - 2(\angle A+\angle B+\angle C) = 180^\circ.$

Location

Up until now, we have been considering an acute triangle. What happens if the triangle is right, or obtuse?

• In an acute triangle, the orthocenter is inside the triangle.
• In a right triangle, the orthocenter is the right-angle vertex.
• In an obtuse triangle, the orthocenter lies outside the triangle.

In an acute triangle, all three altitudes lie inside the triangle. Therefore, the orthocenter also lies inside the triangle.

If $\angle C=90^\circ$, then side $AC$ is perpendicular to $BC$. Since each leg already serves as an altitude, the orthocenter is the right-angle vertex itself.

When a triangle contains an angle greater than $90^\circ$, some altitudes intersect extensions of the sides rather than the sides themselves. As a result, the orthocenter lies outside the triangle.

Worked Example

Triangle $ABC$ has area $24$ and base $BC=8$. Find the altitude from $A$.

Using

we get

Thus

so

More Examples

Example 1: Right Triangle Orthocenter

In a right triangle $ABC$ with right angle at $C$, where is the orthocenter?

Since the legs are already altitudes, the orthocenter is

Example 2: Area and Alternate Base

Triangle $ABC$ has area $30$ and side $AC=10$. Find the altitude from $B$.

Using

we obtain

Thus

Example 3: Orthocenter Location

A triangle has angles

Where is its orthocenter?

Because the triangle is right, the orthocenter lies at the right-angle vertex.

Example 4: Altitude to the Hypotenuse

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into segments of lengths $4$ and $9$. Find the altitude.

Using

we get

Therefore,

Example 5: Triangle DIH

Let $ABC$ be an acute triangle with orthocenter $H$. Let $D$ and $I$ be the feet of the altitudes from $A$ and $B$, respectively.

Find $\angle DIH$.

Since $I$ lies on altitude $BI$, we have

Also, $HD$ lies on altitude $AD$, so

Because $DI$ lies along the side connecting two feet of altitudes, angle chasing shows that $D$, $I$, $B$, and $H$ lie on a common circle.

Therefore,

Thus,

Strategy Checklist

• Mark which segment is the base for the altitude you need.
• Use $A=\tfrac{1}{2}bh$ before introducing extra variables.
• Decide whether the triangle is acute, right, or obtuse before placing the orthocenter.
• Look for right triangles created by altitudes.
• Check for similar triangles whenever an altitude is dropped.
• In right triangles, remember the altitude-to-hypotenuse relationships.
• When multiple altitudes appear, search for cyclic quadrilaterals.

Common Pitfalls

• Forgetting the factor of $\frac{1}{2}$ in the area formula.
• Confusing an altitude with a median.
• Using the wrong base for a given altitude.
• Assuming the orthocenter is always inside the triangle.
• Missing similar triangles created by an altitude.
• Forgetting that an altitude may land on an extension of a side in an obtuse triangle.
• Overlooking cyclic quadrilaterals formed by altitude feet and the orthocenter.

Practice Problems