Geometry

Orthocenter and the Euler line

The three altitudes of a triangle have no obvious reason to agree. Why do they always pass through one point, and where does that point sit?

You need

CentroidGeometry

Where you are 4 in Triangle centers

Geometry11

Triangles18

Triangle centers4

Circumcenter Incenter Centroid

Orthocenter and the Euler line

Why this exists

A triangle has several families of three special lines, and each family that meets in one point does so for a reason you can see. A point on the perpendicular bisector of a side is equally far from the two ends of that side, so where two perpendicular bisectors cross, the point is equally far from all three vertices, and it must lie on the third bisector as well. That is the Circumcenter. The angle bisectors meet for the same kind of reason, with distances to the sides in place of distances to the vertices, at the Incenter. The medians meet at the Centroid, two thirds of the way down each one.

The altitudes are another family of three: from each vertex, the line that meets the opposite side at a right angle. Here no such reason is in sight. An altitude is not the set of points equally far from anything, and in an obtuse triangle two of the altitudes do not even touch the triangle except at their vertex. Yet draw the three altitude lines of any triangle carefully and they pass through one point, every time.

This lesson shows why, works out where that point sits, and finds something less expected still: the point where the altitudes meet, the centroid and the circumcenter always lie on one straight line, with the centroid exactly a third of the way from one end.

Definition

The altitude from a vertex, as defined in Medians and altitudes, is the segment from that vertex perpendicular to the line containing the opposite side; it ends at its foot on that line. In an obtuse triangle the altitudes from the two acute vertices fall outside the triangle and do not cross each other at all, so this lesson works with the whole lines rather than the segments.

The altitude line from A in \triangle ABC is the line through A perpendicular to line BC. The altitude lines from B and C are defined the same way.

The orthocenter of a triangle is the point where its three altitude lines meet. It is written H. (Greek orthos means upright or straight, as in “orthogonal”.)

On the coordinate plane, the altitude line from A passes through A and has the slope that is the negative reciprocal of the slope of \overline{BC}, as in Slopes of perpendicular lines. If \overline{BC} is horizontal, the altitude line from A is the vertical line through A, and if \overline{BC} is vertical, it is horizontal.

Three of a triangle’s centers get a letter each in this lesson: the circumcenter O, the centroid G, and the orthocenter H. In every triangle that is not equilateral these are three different points on one line, and that line is the triangle’s Euler line, named after Leonhard Euler. In an equilateral triangle the three points are the same point, so they do not pick out a line, and an equilateral triangle has no Euler line.

Intuition

Figure 1The three altitudes of an acute triangle cross at one point, the orthocenter H.

Three plumb lines. Stand the triangle on side \overline{BC} and hang a string from A: it falls straight down onto \overline{BC} along the altitude. Now stand it on \overline{CA} and hang a string from B, and then on \overline{AB} and hang one from C. Draw the three strings on one copy of the triangle and they cross at a single point, H, as in Figure 1.

Where it goes depends on the largest angle. In an acute triangle all three altitudes are inside, and so is H. In a right triangle the two legs are themselves altitudes (each leg is perpendicular to the other), so two altitude lines cross at the right-angle vertex, and H is that vertex. In an obtuse triangle the altitude lines from the two acute vertices run outside, and H ends up outside too. Figure 2 shows all three.

Figure 2Acute, right, obtuse. H is inside, then at the right-angle vertex, then outside. Dotted lines are sides extended.

The altitudes are perpendicular bisectors in disguise. Through each vertex draw the line parallel to the opposite side. The three new lines make a triangle twice as large, upside down, and each vertex of the small triangle turns out to be the midpoint of a side of the big one. An altitude of the small triangle passes through such a midpoint and is perpendicular to the big side there (the big side is parallel to the small one). So the altitudes of the small triangle are the perpendicular bisectors of the big one, and perpendicular bisectors are lines we already know meet. The orthocenter of the small triangle is the circumcenter of the big one.

Figure 3Each altitude of \triangle ABC is a perpendicular bisector of the big triangle, so the three meet at the big triangle’s circumcenter.

Results

The altitude lines meet. In every triangle, the three altitude lines pass through one point, the orthocenter H.

Where it sits.

The triangle	Where H is
acute	inside the triangle
right	at the vertex of the right angle
obtuse	outside the triangle

These three answers are all different, so the table can be read backwards too: where H lands tells you whether the triangle is acute, right or obtuse.

Angles at the orthocenter. In an acute triangle ABC,

m\angle BHC = 180^\circ - m\angle A,

and in the same way m\angle CHA = 180^\circ - m\angle B and m\angle AHB = 180^\circ - m\angle C. Also m\angle HBC = 90^\circ - m\angle C and m\angle HCB = 90^\circ - m\angle B.

The Euler line. In every triangle, the centroid G lies on segment \overline{OH} and splits it so that

HG = 2\,GO.

In coordinates this says H = 3G - 2O. Since G is the average of the three vertices, that is

\begin{aligned} x_H &= x_A + x_B + x_C - 2x_O, \\ y_H &= y_A + y_B + y_C - 2y_O. \end{aligned}

In particular, when the circumcenter is the origin, H is found by adding up the vertices.

When the line exists. If any two of O, G and H are the same point, all three are, and this happens exactly when the triangle is equilateral. In every other triangle the three points are different and lie on exactly one line, the Euler line.

Why it is true

The altitude lines meet. Take \triangle ABC. Through A draw the line parallel to \overline{BC}, through B the line parallel to \overline{CA}, and through C the line parallel to \overline{AB}. No two of these are parallel, because no two sides of the triangle are, so they cross in pairs. Call A' the point where the lines through B and C cross, B' the point where the lines through C and A cross, and C' the point where the lines through A and B cross (Figure 3).

In quadrilateral ABCB' both pairs of opposite sides are parallel (\overline{AB'} \parallel \overline{BC} and \overline{CB'} \parallel \overline{AB}), so it is a parallelogram, and opposite sides of a parallelogram are equal (Properties of a parallelogram): AB' = BC. In the same way AC'BC is a parallelogram and AC' = BC. Going from A to B' is the same move as going from B to C, and going from A to C' is the same move as going from C to B, so B' and C' lie on the line through A on opposite sides of A, each a distance BC away. So A is the midpoint of \overline{B'C'}, and in the same way B is the midpoint of \overline{C'A'} and C the midpoint of \overline{A'B'}.

The altitude line from A is perpendicular to \overline{BC}, and \overline{B'C'} is parallel to \overline{BC}, so the altitude line from A is perpendicular to \overline{B'C'} and passes through its midpoint A. That makes it the perpendicular bisector of \overline{B'C'}. In the same way the other two altitude lines of \triangle ABC are the perpendicular bisectors of the other two sides of \triangle A'B'C'. The perpendicular bisectors of any triangle meet at one point, its Circumcenter, so the three altitude lines of \triangle ABC meet at one point, and that point is the circumcenter of \triangle A'B'C'. \blacksquare

Where it sits. Each case uses the table in Medians and altitudes of where an altitude’s foot lands.

Right triangle. Say the right angle is at C. Then \overline{AC} \perp \overline{BC}, so line AC passes through A and is perpendicular to \overline{BC}: it is the altitude line from A. In the same way line BC is the altitude line from B. These two lines cross at C, so H = C.

Acute triangle. Let D be the foot of the altitude from A and E the foot of the altitude from B. In an acute triangle every foot lies strictly between the other two vertices, so D is strictly between B and C and E strictly between A and C. Line AD cuts the triangle into \triangle ABD and \triangle ADC. The point B lies on one side of line AD, and E, being on \overline{AC} but not at A, lies on the side of C. So segment \overline{BE} crosses line AD at a point strictly between B and E, and every such point is inside the triangle. That crossing point is on the altitude lines from A and from B, so it is H, and H is inside.

Obtuse triangle. Say the obtuse angle is at C. The foot D of the altitude from A lands on line BC beyond C, so B, C and D lie on one line in that order. Then B and C are on the same side of D, so the whole of \overline{BC}, and with it the whole triangle except the point A, lies on one side of the altitude line AD. That line touches the triangle only at A. The orthocenter lies on that line and is not A (the altitude line from B would pass through A only if \angle A were right), so H is outside the triangle.

Angles at the orthocenter. Let the triangle be acute, with E the foot of the altitude from B and F the foot of the altitude from C. From the acute case above, H lies strictly between B and E, and in the same way strictly between C and F. The quadrilateral AFHE (Figure 4) has a right angle at E and at F and the angle m\angle A at A. The angles of a quadrilateral add to 360^\circ, so

\begin{aligned} m\angle FHE &= 360^\circ - 90^\circ - 90^\circ - m\angle A \\ &= 180^\circ - m\angle A. \end{aligned}

Since B, H, E lie on one line and C, H, F on another, \angle BHC and \angle FHE are vertical angles, so they are equal, and m\angle BHC = 180^\circ - m\angle A. For the angle at B: triangle BEC has a right angle at E, so m\angle EBC = 90^\circ - m\angle C, and \angle HBC is the same angle because H is on \overline{BE}. The other statements come from relabeling.

Figure 4Two right angles and a leave 180^\circ - a at H, and the vertical angle \angle BHC is the same size.

The Euler line. Keep the big triangle A'B'C' from the first proof, and let M be the midpoint of \overline{BC}. Quadrilateral ABA'C is a parallelogram (its opposite sides lie on parallel lines), and the diagonals of a parallelogram bisect each other, so M is also the midpoint of \overline{AA'}. The centroid G lies on the median \overline{AM} with AG = \tfrac{2}{3}AM and GM = \tfrac{1}{3}AM. The point A' lies on line AM beyond M with MA' = AM, so

GA' = GM + MA' = \tfrac{1}{3}AM + AM = \tfrac{4}{3}AM = 2\,AG.

So A' is reached by starting at A, passing through G, and going on twice as far. The same holds for B' and C'.

Now do that to every point of the plane: send each point P to the point on line PG, on the other side of G, twice as far from G as P is. This is a dilation centered at G by a factor of 2 together with a half turn about G. Both of those send every line to a parallel line, send the midpoint of a segment to the midpoint of its image, and keep right angles right. The move sends A, B, C to A', B', C', so it sends the midpoint M of \overline{BC} to the midpoint of \overline{B'C'}, which is A. It therefore sends the perpendicular bisector of \overline{BC} (through M, perpendicular to \overline{BC}) to the line through A perpendicular to \overline{B'C'}, and so perpendicular to \overline{BC}: the altitude line from A. The same happens for the other two sides. So the point where the three perpendicular bisectors meet, O, is sent to the point where the three altitude lines meet, H. By the way the move was defined, H lies on line OG on the other side of G from O, and GH = 2\,GO. \blacksquare

In coordinates, going from G to H is the move from O to G done twice, so x_H - x_G = 2(x_G - x_O) and x_H = 3x_G - 2x_O = x_A + x_B + x_C - 2x_O, and the same for y.

When the line exists. From HG = 2\,GO: if G = O then HG = 0, so H = G as well; if H = G then GO = 0, so O = G; and if H = O the segment \overline{OH} has length 0, so G, which lies on it, is that point too. So any two equal forces all three equal.

Suppose all three are one point. The median from A passes through G and through the midpoint M of \overline{BC}. The perpendicular bisector of \overline{BC} passes through O = G and through M. The points G and M are different (G is a third of the way up from M to A), so these are the same line: the median from A is perpendicular to \overline{BC}, and A lies on the perpendicular bisector of \overline{BC}. By the Perpendicular bisectors of a triangle, AB = AC. The same argument from B gives BA = BC, so the triangle is equilateral. Conversely, in an equilateral triangle each median is also an altitude and lies along a perpendicular bisector (the isosceles result in Medians and altitudes), so the three families are the same three lines and O = G = H.

Worked examples

Example 1

An acute triangle with a horizontal side

Find the orthocenter of the triangle with vertices A(2, 6), B(-1, 0) and C(7, 0), and say where it lies.

Solution

Two altitude lines are enough to find the point where all three meet, and the third then serves as a check. Start with the easiest. Side \overline{BC} lies on the x-axis, so it is horizontal, and the altitude line from A is the vertical line through A:

x = 2.

For the altitude line from B we need the slope of the side it is perpendicular to:

m_{AC} = \frac{0 - 6}{7 - 2} = -\frac{6}{5}.

Perpendicular slopes multiply to -1, so the altitude line from B has slope \tfrac{5}{6}, and it passes through B(-1, 0):

y = \tfrac{5}{6}(x + 1).

On x = 2 this gives y = \tfrac{5}{6} \cdot 3 = \tfrac{5}{2}, so H = \left(2, \tfrac{5}{2}\right).

Check with the third altitude. \displaystyle m_{AB} = \frac{6 - 0}{2 - (-1)} = 2, so the altitude line from C has slope -\tfrac{1}{2} and equation y = -\tfrac{1}{2}(x - 7). At x = 2 it gives y = -\tfrac{1}{2}(-5) = \tfrac{5}{2}, so it passes through H as well.

H lies on the altitude from A, strictly between A(2, 6) and its foot (2, 0) on \overline{BC}, so H is inside the triangle. Read backwards, the table says the triangle is acute.

Example 2

Two slanted altitudes

Find the orthocenter of the triangle with vertices A(2, 7), B(-3, 2) and C(6, -1).

Solution

No side is horizontal or vertical, so each altitude line needs a slope.

The side opposite A has slope \displaystyle m_{BC} = \frac{-1 - 2}{6 - (-3)} = -\frac{1}{3}, so the altitude line from A has slope 3. Through A(2, 7) it is y - 7 = 3(x - 2), that is, y = 3x + 1.

The side opposite B has slope \displaystyle m_{AC} = \frac{-1 - 7}{6 - 2} = -2, so the altitude line from B has slope \tfrac{1}{2}. Through B(-3, 2) it is y - 2 = \tfrac{1}{2}(x + 3), that is, y = \tfrac{1}{2}x + \tfrac{7}{2}.

H is on both lines, so set the two expressions for y equal:

\begin{aligned} 3x + 1 &= \tfrac{1}{2}x + \tfrac{7}{2} \\ 6x + 2 &= x + 7 \\ x &= 1 \end{aligned}

Then y = 3(1) + 1 = 4, and H = (1, 4).

Check. \displaystyle m_{AB} = \frac{7 - 2}{2 - (-3)} = 1, so the altitude line from C has slope -1: y + 1 = -(x - 6), or y = -x + 5. At x = 1 it gives y = 4. All three altitude lines pass through (1, 4).

Example 3

An obtuse triangle: the orthocenter outside

Find the orthocenter of the triangle with vertices A(0, 0), B(8, 0) and C(2, 3). Show that it lies outside the triangle, and say what that tells you about the triangle.

Figure 5Example 3. The orthocenter H(2, 4) is outside, one unit above the obtuse vertex C.

Solution

Side \overline{AB} is on the x-axis, so the altitude line from C is vertical: x = 2. Its foot is D(2, 0).

The side opposite A has slope \displaystyle m_{BC} = \frac{3 - 0}{2 - 8} = -\frac{1}{2}, so the altitude line from A has slope 2 and passes through the origin: y = 2x. On x = 2 it gives y = 4, so H = (2, 4).

Check. m_{AC} = \tfrac{3}{2}, so the altitude line from B has slope -\tfrac{2}{3}: y = -\tfrac{2}{3}(x - 8). At x = 2, y = -\tfrac{2}{3}(-6) = 4.

Every point of the triangle has y between 0 and 3, the highest point being C, and H has y = 4. So H is outside, one unit straight above C. The table read backwards says the triangle is obtuse. To see which angle: the foot D(2, 0) lies strictly between A and B, and by the table in Medians and altitudes that happens only when the angles at A and B are both acute. So the obtuse angle is at C.

Example 4

Angles at the orthocenter

In acute \triangle ABC, m\angle A = 64^\circ and m\angle B = 71^\circ, and H is the orthocenter. Find m\angle BHC, m\angle CHA, m\angle AHB, m\angle HBC and m\angle HCB.

Solution

The three angles of a triangle add to 180^\circ, so m\angle C = 180^\circ - 64^\circ - 71^\circ, giving m\angle C = 45^\circ.

Each angle at H is the supplement of the triangle’s angle at the vertex it faces:

\begin{aligned} m\angle BHC &= 180^\circ - 64^\circ = 116^\circ, \\ m\angle CHA &= 180^\circ - 71^\circ = 109^\circ, \\ m\angle AHB &= 180^\circ - 45^\circ = 135^\circ. \end{aligned}

Since H is inside the triangle, these three angles go once round H, and indeed 116 + 109 + 135 = 360.

For the angles at B and C in \triangle HBC:

\begin{aligned} m\angle HBC &= 90^\circ - 45^\circ = 45^\circ, \\ m\angle HCB &= 90^\circ - 71^\circ = 19^\circ. \end{aligned}

Check. In \triangle HBC, 116 + 45 + 19 = 180.

Example 5

Finding a vertex from the orthocenter

A triangle has vertices B(-4, 0) and C(6, 0), and its orthocenter is H(2, 3). Find the third vertex A.

Solution

The altitude line from A is perpendicular to the horizontal side \overline{BC}, so it is vertical, and it passes through H. So it is the line x = 2, and A = (2, a) for some number a.

The line through B and H is the altitude line from B, so it is perpendicular to \overline{AC}. Its slope is \displaystyle m_{BH} = \frac{3 - 0}{2 - (-4)} = \frac{1}{2}, so \overline{AC} must have slope -2:

\begin{aligned} \frac{0 - a}{6 - 2} &= -2 \\ -a &= -8 \\ a &= 8 \end{aligned}

So A = (2, 8).

Check with the altitude from C, which was not used: \displaystyle m_{CH} = \frac{3 - 0}{2 - 6} = -\frac{3}{4} and \displaystyle m_{AB} = \frac{8 - 0}{2 - (-4)} = \frac{4}{3}. The product is -1, so line CH is perpendicular to \overline{AB}, as the altitude line from C must be.

Example 6

The Euler line in coordinates

For the triangle of Example 2, with A(2, 7), B(-3, 2), C(6, -1) and H(1, 4), find the circumcenter O and the centroid G. Show that O, G and H lie on one line with HG = 2\,GO, and find the equation of that line.

Figure 6The Euler line of Example 6. G lies between H and O, with HG = 2\,GO.

Solution

The circumcenter is where perpendicular bisectors cross. Side \overline{AB} has midpoint \left(-\tfrac{1}{2}, \tfrac{9}{2}\right) and slope 1, so its perpendicular bisector has slope -1: y - \tfrac{9}{2} = -\left(x + \tfrac{1}{2}\right), or y = -x + 4. Side \overline{BC} has midpoint \left(\tfrac{3}{2}, \tfrac{1}{2}\right) and slope -\tfrac{1}{3}, so its perpendicular bisector has slope 3: y - \tfrac{1}{2} = 3\left(x - \tfrac{3}{2}\right), or y = 3x - 4. Setting -x + 4 = 3x - 4 gives x = 2 and y = 2, so O(2, 2). As a check, O is 5 from each vertex: OA^2 = 0 + 25, OB^2 = 25 + 0, OC^2 = 16 + 9.

The centroid is the average of the vertices:

\begin{aligned} G &= \left(\frac{2 - 3 + 6}{3}, \frac{7 + 2 - 1}{3}\right) \\ &= \left(\tfrac{5}{3}, \tfrac{8}{3}\right). \end{aligned}

So G\left(\tfrac{5}{3}, \tfrac{8}{3}\right).

Compare the move from O to G with the move from G to H:

\begin{aligned} G - O &= \left(-\tfrac{1}{3}, \tfrac{2}{3}\right), \\ H - G &= \left(-\tfrac{2}{3}, \tfrac{4}{3}\right). \end{aligned}

The second move is exactly twice the first: the same direction, twice as far. So O, G, H lie on one line, with G between O and H and HG = 2\,GO. In lengths, \displaystyle GO = \sqrt{\tfrac{1}{9} + \tfrac{4}{9}}, so \displaystyle GO = \tfrac{\sqrt{5}}{3}, and \displaystyle HG = \sqrt{\tfrac{4}{9} + \tfrac{16}{9}}, so \displaystyle HG = \tfrac{2\sqrt{5}}{3}, twice as long.

The line through O(2, 2) has slope \displaystyle \frac{2/3}{-1/3} = -2: y - 2 = -2(x - 2), which is y = -2x + 6. Check: H(1, 4) gives -2 + 6 = 4, and G gives -\tfrac{10}{3} + \tfrac{18}{3} = \tfrac{8}{3}.

Example 7

The orthocenter without altitudes

Show that the circumcenter of the triangle with vertices A(3, 4), B(-5, 0) and C(4, -3) is the origin, use the Euler line to find its orthocenter, and confirm the answer with two altitudes.

Solution

Each vertex is 5 from the origin: 3^2 + 4^2 = 25, (-5)^2 + 0^2 = 25 and 4^2 + (-3)^2 = 25. The circumcenter is the only point equally far from all three vertices, so it is O(0, 0).

With O at the origin, H = 3G - 2O is just the sum of the vertices:

H = (3 - 5 + 4, \; 4 + 0 - 3) = (2, 1).

So H = (2, 1).

Confirm. \displaystyle m_{BC} = \frac{-3 - 0}{4 - (-5)} = -\frac{1}{3}, so the altitude line from A has slope 3: y - 4 = 3(x - 3), or y = 3x - 5, and at x = 2 it gives y = 1. Next, \displaystyle m_{AC} = \frac{-3 - 4}{4 - 3} = -7, so the altitude line from B has slope \tfrac{1}{7}: y = \tfrac{1}{7}(x + 5), and at x = 2 it gives y = 1. Both pass through (2, 1). The Euler line did in one line of arithmetic what two altitudes and a pair of equations did the long way.

Boundaries

Three points on one line have no orthocenter. If A, B and C are collinear, every “altitude line” is perpendicular to that same line, so the three are parallel and never meet. Everything in this lesson needs a genuine triangle.

It is the lines that meet, not the segments. In an obtuse triangle the altitude segments from the two acute vertices end at their feet, outside the triangle, and do not reach H. Only the full lines pass through it. In a right triangle two of the altitude lines are sides.

The orthocenter measures nothing. The circumcenter is equally far from the three vertices and the incenter is equally far from the three sides, which is why each is the center of a circle. The orthocenter has no such property. In Example 1, HA = 6 - \tfrac{5}{2} = \tfrac{7}{2} but \displaystyle HB = \sqrt{3^2 + \left(\tfrac{5}{2}\right)^2} = \tfrac{\sqrt{61}}{2}, which is not \tfrac{7}{2}.

The angle relation was proved for acute triangles. The argument used the quadrilateral AFHE with H inside the triangle. When H is outside, the quadrilateral is a different shape and the angles at H have to be worked out again from the picture.

The Euler line needs a triangle that is not equilateral. In an equilateral triangle O, G and H are one point and determine no line. The ratio HG = 2\,GO holds in every other triangle, even when H and O are both outside it. The fourth classic center, the incenter, is not on the Euler line in general; it lies on it when the triangle is isosceles but not equilateral, since then every center is on the line of symmetry.

Where this goes next

The orthocenter completes the four classic Triangle centers: circumcenter, incenter, centroid and orthocenter, three of which line up on the Euler line.

The feet of the three altitudes are the vertices of a smaller triangle inside an acute triangle, called the orthic triangle. In Figure 4 the quadrilateral AFHE has two opposite right angles, so its opposite angles add to 180^\circ and its four vertices lie on one circle: it is a Inscribed polygons and cyclic quadrilaterals. Facts like this are where the study of triangle geometry goes after the centers. The best known is that the midpoints of the three sides, the three feet of the altitudes and the midpoints of \overline{HA}, \overline{HB} and \overline{HC} all lie on one circle, the nine-point circle, whose center is the midpoint of \overline{OH}, on the Euler line.

The formula H = A + B + C when O is at the origin is a statement about adding points as arrows, and it is the natural way into proving results like these with Vectors.

Exercises

Finding the orthocenter

Find the orthocenter of the triangle with vertices A(4, 8), B(0, 0) and C(10, 0).
Find the orthocenter of the triangle with vertices A(-3, 4), B(-4, -3) and C(5, 0).
Find the orthocenter of the triangle with vertices A(0, 0), B(6, 0) and C(-2, 4). Is it inside or outside the triangle, and which angle of the triangle is obtuse?
The triangle has vertices P(1, 2), Q(5, 4) and R(3, 8). Show that it has a right angle, and use that to give its orthocenter with no further work.
Find the orthocenter of the triangle with vertices A(1, 1), B(7, 3) and C(3, 7), and check it with the third altitude.

Where the orthocenter sits

Where is the orthocenter of a triangle whose angles measure (a) 50^\circ, 60^\circ, 70^\circ; (b) 30^\circ, 60^\circ, 90^\circ; (c) 20^\circ, 35^\circ, 125^\circ?
(a) A triangle has angles (2x)^\circ, (3x)^\circ and (4x)^\circ. Where is its orthocenter? (b) Answer the same question for angles x^\circ, x^\circ and (4x)^\circ.

Angles at the orthocenter

In acute \triangle ABC, m\angle A = 50^\circ, m\angle B = 60^\circ and m\angle C = 70^\circ. Find m\angle BHC, m\angle CHA and m\angle AHB.
In acute \triangle ABC with orthocenter H, m\angle BHC = 128^\circ. Find m\angle A.
In acute \triangle ABC, m\angle A = 58^\circ and m\angle B = 74^\circ. Find m\angle HBC and m\angle HCB, and check them with the angle sum of \triangle HBC.
In an acute triangle, m\angle AHB is three times m\angle C. Find m\angle C.

Working backwards

A triangle has vertices B(0, 0) and C(9, 0), and its orthocenter is H(3, 2). Find the third vertex A.
Show that the triangle with vertices A(4, 3), B(-3, 4) and C(0, -5) has its circumcenter at the origin. Use the Euler line to find its orthocenter, and confirm it with one altitude.
A triangle has circumcenter O(1, 2) and centroid G(3, 1). Find its orthocenter.
A triangle has orthocenter H(-4, 9) and centroid G(2, 3). Find its circumcenter.

The Euler line

The triangle has vertices C(0, 0), A(0, 9) and B(12, 0). Find H, G and O, find the equation of the Euler line, and check that HG = 2\,GO.
The triangle has vertices A(0, 6), B(-4, 0) and C(4, 0). Find H, G and O. What is the Euler line?
For the triangle of Example 3, with A(0, 0), B(8, 0) and C(2, 3), find G and O, and check that 3G - 2O is the orthocenter found there.

Reasoning

The orthocenter H of \triangle ABC is not one of its vertices. Show that A is the orthocenter of \triangle HBC.
Explain why the orthocenter of a right triangle is the vertex of the right angle, and why its Euler line is the median from that vertex.
Prove, without using the Euler line, that if the orthocenter and the centroid of a triangle are the same point, the triangle is equilateral.
Show that the orthocenter can lie on the line through two of the vertices only if it is one of the vertices.
In \triangle ABC, AB = AC, and the triangle is not equilateral. Show that its Euler line is the line through A and the midpoint of \overline{BC}.

Answers

H(4, 3). Side \overline{BC} is horizontal, so the altitude line from A is x = 4. \displaystyle m_{AC} = \frac{0 - 8}{10 - 4} = -\frac{4}{3}, so the altitude line from B is y = \tfrac{3}{4}x, which gives y = 3 at x = 4. Check: m_{AB} = 2, so the altitude line from C is y = -\tfrac{1}{2}(x - 10), which also gives 3 at x = 4.
H(-2, 1). \displaystyle m_{BC} = \frac{0 + 3}{5 + 4} = \frac{1}{3}, so the altitude line from A has slope -3: y = -3x - 5. \displaystyle m_{AC} = \frac{0 - 4}{5 + 3} = -\frac{1}{2}, so the altitude line from B has slope 2: y = 2x + 5. Then -3x - 5 = 2x + 5 gives x = -2 and y = 1.
H(-2, -4), outside, and the obtuse angle is at A. The altitude line from C is x = -2. m_{AC} = -2, so the altitude line from B is y = \tfrac{1}{2}(x - 6), which gives y = -4 at x = -2. The triangle lies on or above the x-axis and H is below it, so H is outside, and the triangle is obtuse. The foot (-2, 0) of the altitude from C lies on line AB beyond A, which by the table in Medians and altitudes means \angle A is the obtuse angle.
H = Q(5, 4). \displaystyle m_{QP} = \frac{2 - 4}{1 - 5} = \frac{1}{2} and \displaystyle m_{QR} = \frac{8 - 4}{3 - 5} = -2. The product is -1, so \angle Q is a right angle, and the orthocenter of a right triangle is the vertex of the right angle.
H(4, 4). m_{BC} = -1, so the altitude line from A is y = x. m_{AC} = 3, so the altitude line from B is y = -\tfrac{1}{3}x + \tfrac{16}{3}. Then x = -\tfrac{1}{3}x + \tfrac{16}{3} gives x = 4. Check: m_{AB} = \tfrac{1}{3}, so the altitude line from C is y = -3x + 16, and -3(4) + 16 = 4.
(a) Inside, since all three angles are acute. (b) At the vertex with the 90^\circ angle. (c) Outside, since 125^\circ is obtuse.
(a) 2x + 3x + 4x = 180 gives x = 20, so the angles are 40^\circ, 60^\circ and 80^\circ, all acute: the orthocenter is inside. (b) x + x + 4x = 180 gives x = 30, so the angles are 30^\circ, 30^\circ and 120^\circ: the orthocenter is outside.
m\angle BHC = 180^\circ - 50^\circ = 130^\circ, m\angle CHA = 180^\circ - 60^\circ = 120^\circ and m\angle AHB = 180^\circ - 70^\circ = 110^\circ. They add to 360^\circ, as angles once round a point must.
m\angle A = 52^\circ, from 128 = 180 - m\angle A.
m\angle C = 180^\circ - 58^\circ - 74^\circ = 48^\circ. Then m\angle HBC = 90^\circ - 48^\circ = 42^\circ and m\angle HCB = 90^\circ - 74^\circ = 16^\circ. Check: m\angle BHC = 180^\circ - 58^\circ = 122^\circ, and 42 + 16 + 122 = 180.
m\angle C = 45^\circ. Write c for m\angle C; then m\angle AHB = 180 - c, and 180 - c = 3c gives c = 45.
A(3, 9). The altitude line from A is vertical through H, so A = (3, a). Line BH has slope \tfrac{2}{3} and is perpendicular to \overline{AC}, so \displaystyle \frac{0 - a}{9 - 3} = -\frac{3}{2} and a = 9. Check: m_{CH} = -\tfrac{1}{3} and m_{AB} = 3, product -1.
H(1, 2). Each vertex is 5 from the origin (16 + 9 = 9 + 16 = 0 + 25 = 25), so O(0, 0), and H = A + B + C = (4 - 3 + 0, 3 + 4 - 5). Confirm: m_{AB} = -\tfrac{1}{7}, so the altitude line from C is y = 7x - 5, which passes through (1, 2).
H(7, -1), from H = 3G - 2O = (9 - 2, 3 - 4).
O(5, 0). From H = 3G - 2O, 2O = 3G - H = (6 + 4, 9 - 9) = (10, 0).
H(0, 0), G(4, 3), O\left(6, \tfrac{9}{2}\right), Euler line y = \tfrac{3}{4}x. The right angle is at C, so H = C. G is the average of the vertices, and O is the midpoint of the hypotenuse \overline{AB} (Circumcenter). The line through (0, 0) and (4, 3) has slope \tfrac{3}{4}, and O is on it since \tfrac{3}{4} \cdot 6 = \tfrac{9}{2}. Then HG = 5 and \displaystyle GO = \sqrt{2^2 + \left(\tfrac{3}{2}\right)^2} = \tfrac{5}{2}, so HG = 2\,GO.
H\left(0, \tfrac{8}{3}\right), G(0, 2), O\left(0, \tfrac{5}{3}\right), and the Euler line is x = 0, the line of symmetry. The altitude line from A is x = 0; m_{AC} = -\tfrac{3}{2}, so the altitude line from B is y = \tfrac{2}{3}(x + 4), giving \tfrac{8}{3} at x = 0. For O(0, k): 16 + k^2 = (6 - k)^2 gives k = \tfrac{5}{3}.
G\left(\tfrac{10}{3}, 1\right) and O\left(4, -\tfrac{1}{2}\right). O is on x = 4, the perpendicular bisector of \overline{AB}, and OA^2 = OC^2 gives 16 + y^2 = 4 + (y - 3)^2, so y = -\tfrac{1}{2}. Then 3G - 2O = (10 - 8, 3 + 1) = (2, 4), which is H.
By Exercise 22, H is not on line BC, so \triangle HBC is a triangle. Line AH is the altitude line of \triangle ABC from A, so it is perpendicular to \overline{BC}; it passes through H, so it is the altitude line of \triangle HBC from H. Line BH is the altitude line from B, so \overline{BH} \perp \overline{AC}; then line CA passes through C and is perpendicular to \overline{BH}, so it is the altitude line of \triangle HBC from C. In the same way line BA passes through B and is perpendicular to \overline{CH}, so it is the altitude line from B. All three pass through A, so A is the orthocenter of \triangle HBC.
With the right angle at C, each leg is perpendicular to the other, so line AC is the altitude line from A and line BC is the altitude line from B, and they cross at C: H = C. The circumcenter of a right triangle is the midpoint of the hypotenuse (Circumcenter). A right triangle is not equilateral, so its Euler line is the line through H = C and O, the midpoint of \overline{AB}, which is the median from C.
Suppose H = G. The altitude line from A and the median from A both pass through A and through G, and G \ne A because G is inside the triangle. Two different points lie on only one line, so the median \overline{AM} to the midpoint M of \overline{BC} lies along the altitude and is perpendicular to \overline{BC}. Then line AM is the perpendicular bisector of \overline{BC}, and A is on it, so AB = AC. The same argument from B gives BA = BC, so all three sides are equal.
Suppose H is on line BC and H \ne B. The altitude line from B passes through B and H, so it is line BC. That line is perpendicular to \overline{AC}, so \angle C is a right angle, and then H = C by Exercise 20. So H is B or C. The same holds for the other two sides.
Let M be the midpoint of \overline{BC}. Since AB = AC, the median \overline{AM} is also the altitude from A and lies on the perpendicular bisector of \overline{BC} (the isosceles result in Medians and altitudes). So line AM contains G (it is a median), H (it is an altitude line) and O (it is a perpendicular bisector). The triangle is not equilateral, so O, G, H are three different points, and the one line through them is line AM.