Altitudes and the Orthocenter

Watch the video explainer

Learning goals

Construct triangle altitudes using the full supporting lines of opposite sides
Prove altitude concurrence by identifying perpendicular bisectors of an auxiliary triangle
Locate the orthocenter in acute, right, and obtuse triangles with geometric reasons
Calculate lengths using the right triangles created by an altitude
Compare four triangle centers and justify their coincidence in an equilateral triangle

A perpendicular to the opposite side line

An altitude of a nondegenerate triangle is a perpendicular segment from a vertex to the line containing the opposite side. Its endpoint on that line is the foot of the altitude. Its length is the perpendicular distance from the vertex to the line.

For example, a 5 cm perpendicular from $A$ to line $BC$ is an altitude of length 5 cm. A 7 cm slanted segment from $A$ to that same line is not an altitude. The right angle, rather than the size of the length, is the defining condition.

AD is an altitude because it joins vertex A perpendicularly to the opposite side line.

The altitude line is the full line containing an altitude. We use these full lines when discussing their common intersection. In an obtuse triangle, some altitude segments lie outside the triangle, and their lines may need extending further to reach the common point.

Construct the foot, including an extension

Use the perpendicular-from-a-point construction from the first lesson of this chapter. Draw the complete line through the opposite side before deciding where its foot will fall. The paper’s horizontal direction has no role in the definition.

Worked example 1 Construct an exterior altitude

Problem. Triangle $ABC$ has an obtuse angle at $A$ . Construct the altitude from $B$ with a compass and unmarked straightedge. Identify its foot $E$ and explain why a side extension is needed.

Solution. Extend line $AC$ through $A$ . Draw a circle centered at $B$ with radius greater than $BA$ . Since $A$ lies inside this circle, line $AC$ cuts it at two distinct points, $U$ and $V$ .

Construct the perpendicular bisector of $UV$ . The equal radii $BU$ and $BV$ put $B$ on this bisector, so it is the perpendicular from $B$ to line $AC$ . Its intersection with $AC$ is $E$ , and $BE$ is the required altitude.

The foot lies beyond $A$ on the extension. If $E$ were on ray $AC$ , the right triangle $BAE$ would have the obtuse angle $BAC$ at $A$ , which is impossible. The foot cannot be $A$ either, because the angle at $A$ is not right.

Extend AC past A to reach the foot E; the vertex of this altitude is B.

Check your understanding

Which condition identifies the altitude from vertex P in triangle PQR?

A It joins P to the midpoint of QR B It divides the angle at P equally C It passes through P and is perpendicular to line QR D It is parallel to QR

Make a larger triangle to prove concurrence

The three altitude lines of any nondegenerate triangle are concurrent. Their common point is the orthocenter, usually labeled $H$ .

To see why, build a triangle whose perpendicular bisectors are the original altitude lines. Through $A$ draw the parallel to $BC$ , through $B$ the parallel to $CA$ , and through $C$ the parallel to $AB$ . Name their intersections so that $A$ lies on $QR$ , $B$ on $RP$ , and $C$ on $PQ$ .

Stage 1. Each auxiliary side passes through an original vertex and is parallel to the opposite original side.

Why the altitude lines meet at one point#

First take a triangle with $AB$ of 9 units, $AC$ of 10 units, and $BC$ of 12 units. The parallels make parallelograms $ABCQ$ and $ACBR$ . Opposite sides in these give $AQ$ and $AR$ the same length as $BC$ , so each is 12 units. Thus $A$ is the midpoint of the 24-unit auxiliary side $QR$ . The two auxiliary halves at $B$ are each 10 units, and those at $C$ are each 9 units. Thus all three original vertices become midpoints. The altitude through each midpoint is perpendicular to the auxiliary side, so the three lines are the larger triangle’s perpendicular bisectors. Those meet at one point. The lengths 9, 10, and 12 are not needed for any step.

For any nondegenerate $ABC$ , the three original side directions are distinct. Thus the pairwise intersections of the constructed parallels are well defined. To check their order on the line through $A$ , translate $A$ by the directed move from $B$ to $C$ . Its image is $Q$ : it lies on the parallel through $A$ to $BC$ and on the parallel through $C$ to $BA$ . The reverse move puts $R$ on the opposite ray from $A$ . Hence $A$ lies between $Q$ and $R$ . The same argument establishes the corresponding orders at $B$ and $C$ .

The parallel pairs make $ABCQ$ , $ACBR$ , and $ABPC$ parallelograms. Their opposite-side equalities give

AQ=BC.

Also,

AR=BC.

Therefore $A$ is the midpoint of $QR$ . The remaining parallelogram equalities give

RB=AC.

And

BP=AC.

Thus $B$ is the midpoint of $RP$ . Finally,

PC=AB.

And

CQ=AB.

Thus $C$ is the midpoint of $PQ$ . Each side has positive length; in particular $Q$ and $R$ are distinct. The other auxiliary sides have different directions from $QR$ , so $PQR$ is a nondegenerate triangle.

Stage 2. The parallelograms prove the three midpoint pairs. Different tick patterns refer to different equalities.

The altitude line from $A$ is perpendicular to $BC$ . Since $QR$ is parallel to $BC$ , this same line is perpendicular to $QR$ . It passes through its midpoint $A$ , so it is the perpendicular bisector of $QR$ .

Likewise, the altitude line from $B$ is the perpendicular bisector of $RP$ , and the altitude line from $C$ is the perpendicular bisector of $PQ$ .

The perpendicular-bisector concurrence theorem applies to $PQR$ . Its unique circumcenter lies on all three original altitude lines. This proves both their concurrence and the uniqueness of their common point.

Stage 3. The same lines are altitudes of ABC and perpendicular bisectors of the auxiliary triangle PQR.

The orthocenter of $ABC$ is the circumcenter of this auxiliary triangle. The proof does not identify it as the circumcenter of $ABC$ itself.

Start with H and choose acute, right and obtuse cases. Drag A or vary the base angles while BC stays fixed at 6 units. Use Next to inspect O, I and G separately before showing all four centers. Finally choose Equilateral and compare their locations. The drawing illustrates the separate geometric constructions.

Compare the four triangle centers through changing angle cases

Compare orthocenter H

Scroll sideways to see the whole diagram.

Angle B Angle C Show center Try a case

Angle B = 40°. Angle C = 75°. Angle A = 65°. The triangle is acute. The three altitude lines meet at H. H is inside the triangle. Each square mark identifies a perpendicular foot.

Altitude lines include the necessary supporting-line extensions. The final comparison hides helper lines so nearby centers and exact coincidences are easier to distinguish.

Worked example 2 Use the auxiliary midpoint rather than an original midpoint

Problem. In the auxiliary construction above, $BC$ is 14 cm long. Find $AQ$ , $AR$ , and $QR$ . Explain why the altitude line through $A$ bisects $QR$ , even though it need not bisect $BC$ .

Solution. Opposite sides of $ABCQ$ give

AQ=14\text{ cm}.

Opposite sides of $ACBR$ give

AR=14\text{ cm}.

Since $A$ is between $Q$ and $R$ ,

QR=28\text{ cm}.

The midpoint of the auxiliary side is $A$ itself. The altitude line passes through $A$ and is perpendicular to $QR$ . Nothing here establishes that its foot on the original side $BC$ is a midpoint. Stages 2 and 3 show the two roles of the same line.

Check your understanding

What must be established before calling the altitude through A a perpendicular bisector of auxiliary side QR?

A A is the midpoint of QR, and QR is parallel to BC B The original triangle has equal sides C The altitude bisects the original side BC D The original centroid lies on QR

The angle type decides the location

An altitude foot and the orthocenter are different points. The foot is on a supporting side line; the orthocenter is on all three altitude lines.

Worked example 3 Locate the orthocenter in three cases

Problem. Locate the orthocenter of triangle $ABC$ in each case, and give a geometric reason.

Case (a): all three angles are acute.

Case (b): the angle at $C$ is right.

Case (c): the angle at $A$ is obtuse.

Solution. In case (a), the feet lie strictly inside the opposite sides. A foot beyond an endpoint would force a right triangle to have a second angle at least 90 degrees, contradicting an acute base angle. Two vertex-to-opposite-interior-point segments cross inside the triangle. Their crossing is $H$ .

Case (a). The altitude segments themselves cross inside the acute triangle.

In case (b), $AC$ is perpendicular to $BC$ , so the two legs are already altitude segments from $A$ and $B$ . They meet at $C$ . The third altitude starts at $C$ , so $C$ is the orthocenter.

Case (b). Two altitude feet coincide at the right-angle vertex C.

In case (c), the foot $E$ from $B$ lies on $AC$ extended past $A$ , as in Example 1. Thus $A$ and $C$ lie on the same side of line $BE$ ; the line meets the triangle only at $B$ . The altitude through $A$ does not pass through $B$ , because that would make the angle at $B$ right. Their crossing $H$ therefore lies outside the triangle. The altitude from $C$ also has its foot beyond $A$ on the other side line.

Case (c). Distinguish the exterior feet E and F from the exterior intersection H.

The three cases cover every nondegenerate triangle: $H$ is inside for acute triangles, at the right-angle vertex for right triangles, and outside for obtuse triangles.

Check your understanding

A nondegenerate triangle has its orthocenter on its boundary. What follows?

A It is acute B It is obtuse C It is equilateral D It is right, with the orthocenter at the right-angle vertex

An altitude creates useful right triangles

Once a perpendicular is drawn, the earlier right-triangle tools apply. Keep the altitude length separate from the two lengths along the side.

Worked example 4 Find an altitude and the remaining side

Problem. In triangle $ABC$ , point $D$ lies between $B$ and $C$ , and $AD$ is perpendicular to $BC$ . The lengths are

AB=13\text{ cm}.

Also,

BD=5\text{ cm}.

And

DC=9\text{ cm}.

Find $AD$ and $AC$ .

The right angle at D supplies two right triangles sharing the altitude AD.

Solution. In right triangle $ABD$ , $AB$ is the hypotenuse. The Pythagorean theorem gives

\begin{aligned} AD^2&=13^2-5^2\\ AD^2&=144\\ AD&=12\text{ cm}. \end{aligned}

Lengths are positive, so the negative square root is excluded. In right triangle $ACD$ , $AC$ is the hypotenuse.

\begin{aligned} AC^2&=12^2+9^2\\ AC^2&=225\\ AC&=15\text{ cm}. \end{aligned}

Check your understanding

In Example 4, which segment is the hypotenuse of right triangle ACD?

AD

AC

CD

BD

Compare the four centers

Each center is selected by a different construction. A generic altitude has no midpoint or equal-angle promise.

Center	Defining lines	What the point does
Circumcenter	Perpendicular bisectors of the sides	Is equally far from the three vertices
Incenter	Internal angle bisectors	Is inside and equally far from the supporting side lines
Centroid	Medians	Is inside and divides every median two to one from the vertex
Orthocenter	Altitude lines	Is inside, at a vertex, or outside according to the triangle’s angle type

The circumcenter is inside an acute triangle, at the hypotenuse midpoint in a right triangle, and outside an obtuse triangle. Thus a right triangle already separates the orthocenter from the circumcenter: one is its right-angle vertex, the other its hypotenuse midpoint.

Worked example 5 Explain the equilateral coincidence

Problem. Triangle $ABC$ is equilateral, and $D$ is the midpoint of $BC$ . Prove that $AD$ is a median, an internal angle bisector, an altitude, and a perpendicular bisector. Explain why all four triangle centers coincide.

Solution. The midpoint condition makes $AD$ a median. Triangles $ABD$ and $ACD$ have equal side lengths $AB$ and $AC$ , equal side halves $BD$ and $DC$ , and the common side $AD$ . SSS makes them congruent, so the two angle parts at $A$ are equal. Thus $AD$ is the internal angle bisector.

The two angles at $D$ are also equal. They form a straight angle, so each is 90 degrees. Hence $AD$ is an altitude, and the line containing it is also a perpendicular bisector of $BC$ .

Repeat at $B$ and $C$ . The three defining line families are now the same for all four centers, so their unique common point is shared.

In the equilateral case, the same three lines perform all four constructions.

Check your understanding

Which condition guarantees that all four triangle centers coincide?

A The triangle has one obtuse angle B Two altitudes are different lengths C The triangle is equilateral D One side is horizontal

Altitudes and the Orthocenter

Learning goals

A perpendicular to the opposite side line

Construct the foot, including an extension

Make a larger triangle to prove concurrence

The angle type decides the location

An altitude creates useful right triangles

Compare the four centers

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress