Thirteen proofs of the existence of the orthocenter

Introduction

Let us begin with the statement of the orthocenter theorem.

Orthocenter theorem.

The altitudes of triangle \(ABC\) are concurrent.

The orthocenter of triangle \(ABC\) is the point where its altitudes meet — and it is this existence that we will prove. Most proofs start from the intersection point \(H\) of two altitudes \(BE\) and \(CF\) and then verify that the third altitude passes through this point as well, i.e. that the line \(AH\) is perpendicular to \(BC\).

Note that any two altitudes intersect, since no two sides of the triangle are parallel. For simplicity we carry out all arguments for an acute triangle; they remain valid in the obtuse case, except that the intersection point lies outside the triangle. In a right triangle the altitudes meet at the right-angled vertex — a degenerate situation.

First proof — inscribed angles

The simplest proof is, alas, not the most elementary. It requires familiarity with cyclic quadrilaterals. Suppose the altitudes \(BE\) and \(CF\) meet at \(H\). Draw the line \(AH\) until it meets \(BC\) at \(D\), and show that \(\angle ADC\) is a right angle.

Indeed, quadrilateral \(AEHF\) is cyclic with diameter \(AH\), hence \(\angle AFE = \angle AHE\). Also, quadrilateral \(BFEC\) is cyclic with diameter \(BC\), hence \(\angle AFE = \angle ACB\).

Therefore \(\angle AHE = \angle ACB\), so quadrilateral \(CDHE\) is cyclic as well. Since \(\angle HEC = 90^\circ\), we also have \(\angle HDC = 90^\circ\).

Second proof — perpendicular bisectors

This proof is a textbook favorite, though in my view its pedagogical value is somewhat overstated. The auxiliary construction is almost impossible to discover on one’s own, and few problems call for it. Still, the general idea — reduce an unfamiliar claim to a known one — is very useful. When proving that three lines are concurrent, it is often a good strategy to show that they are, for example, the angle bisectors of some triangle, or its perpendicular bisectors, or its medians (being careful not to argue in a circle).

Here is the proof. Through the vertices of \(ABC\) draw lines parallel to the opposite sides. You obtain a larger triangle of which \(ABC\) is the medial triangle (three parallelograms are visible from the construction). Recall that the medial triangle is formed by joining the midpoints of the sides.

The altitudes of \(ABC\) are perpendicular to the sides of the larger triangle, hence they are the perpendicular bisectors of those sides. The perpendicular bisectors of any triangle are concurrent (at the circumcenter).

Third proof — angle bisectors, first approach

One can relate the concurrency of the altitudes not only to perpendicular bisectors but also to angle bisectors. The altitudes of the original triangle are the internal angle bisectors of the orthic triangle, i.e. the triangle whose vertices are the feet of those altitudes.

There are several ways to see this. We begin with a more elementary route. As before, let \(E\) and \(F\) be the feet from \(B\) and \(C\), and let \(M\) be the midpoint of \(BC\). Then \(M\) is the midpoint of the hypotenuse in both right triangles \(BEC\) and \(BFC\). Hence \(MB = MC = MF = ME\), and we have the angle equalities

\[ \angle MFB=\angle B,\quad \angle EMC=\angle C, \quad \angle MFE=\angle MEF. \]

In quadrilateral \(BFEC\) the sum of the angles equals twice the sum of the angles of \(ABC\), so the last equality can be strengthened to

\[ \angle MFE=\angle MEF=\angle A. \]

Using once more that the angle sum of \(ABC\) is \(180^\circ\), we conclude that

\[ \angle AEF=\angle B,\quad \angle AFE=\angle C. \]

Now let \(D\) be the foot from \(A\) (we may ignore \(M\) in the figure). By the same reasoning,

\[ \angle CED=\angle B,\quad \angle BFD=\angle C,\quad \angle BDF=\angle CDE=\angle A, \]

which means that the sides of \(ABC\) bisect the exterior angles of triangle \(DEF\), and the vertices \(A\), \(B\), \(C\) are the excenters. Therefore \(A\), \(B\), and \(C\) lie on the internal bisectors of angles \(D\), \(E\), and \(F\), and the lines \(AD\), \(BE\), and \(CF\) are concurrent.

Third proof — angle bisectors, second approach

The same circle of ideas can be presented differently if you already know about cyclic quadrilaterals. On the one hand, the equalities of the blue, red, and green angles in the figure above follow at once from cyclicity of \(BFEC\), \(AEDB\), and \(CDFA\). On the other hand, the same cyclic facts yield the bisector condition directly, for instance as the chain

\[ \angle FDA = \angle FCA = \angle EBA = \angle EDA. \]

Fourth proof — radical axes

Another way to prove concurrency is to exhibit three circles for which the given lines are radical axes.

The radical axis of two circles is the locus of points having equal power with respect to both circles. For two non-concentric circles it is always a line; if the circles intersect, the radical axis is nothing but the line through their common chord. For three circles whose centers are not collinear, the three pairwise radical axes meet at one point — the radical center. In particular, for three pairwise intersecting circles with non-collinear centers, the three common chords are concurrent as well.

How can we use this to prove that the altitudes meet at one point? Consider the circles with the sides of the triangle as diameters. Their centers are not collinear (they are the midpoints of the sides), and each altitude of the triangle is a common chord of two of these circles!

Fifth proof — Pythagoras

The next proof is a standard choice if you prefer to argue via perpendicularity. As before, let \(H\) be the intersection of the altitudes \(BE\) and \(CF\), and prove that \(AH\) is perpendicular to \(BC\). To show that two segments are perpendicular, the following lemma about quadrilaterals with perpendicular diagonals can help.

Lemma on perpendicular diagonals.

The diagonals of a quadrilateral are perpendicular if and only if the sums of the squares of pairs of opposite sides are equal.

This is an easy consequence of the Pythagorean theorem. Its real advantage is that convexity is irrelevant: for the (possibly non-convex) quadrilateral \(ABHC\), perpendicularity of the diagonals is equivalent to

\[ BH^2 + AC^2 = CH^2 + AB^2. \]

We already know that in quadrilaterals \(BCHA\) and \(CAHB\) the diagonals are perpendicular, so the same criterion gives

\[ AH^2 + BC^2 = BH^2 + AC^2 \quad \text{and} \quad AH^2 + BC^2 = CH^2 + AB^2. \]

Hence \(BH^2 + AC^2 = CH^2 + AB^2\), i.e. the diagonals of \(ABHC\) are perpendicular.

Sixth proof — Carnot’s theorem

A similar argument can be packaged using Carnot’s theorem. One convenient formulation is as follows.

Carnot’s theorem.

Let \(ABC\) be a triangle, and let \(D\), \(E\), and \(F\) be points on the lines \(BC\), \(CA\), and \(AB\), respectively. Through these points draw the lines perpendicular to \(BC\), \(CA\), and \(AB\), respectively. These three perpendiculars are concurrent if and only if \[ BD^2 - DC^2 + CE^2 - EA^2 + AF^2 - FB^2 = 0. \]

In our setting we exploit the abundance of right triangles. In right triangles \(ABD\) and \(ACD\), \[ BD^2+DA^2=BA^2 \quad \text{and} \quad AD^2+DC^2=AC^2. \] Subtracting the second equality from the first, \[ BD^2-DC^2=BA^2-AC^2. \] Similarly, \[ CE^2-EA^2=CB^2-BA^2 \quad \text{and} \quad AF^2-FB^2=AC^2-CB^2. \] Adding the three identities yields \[ BD^2-DC^2 + CE^2-EA^2 +AF^2-FB^2=0. \]

Seventh proof — Ceva’s theorem

One of the most versatile tools for proving that three lines are concurrent is Ceva’s theorem, especially when those lines pass through the vertices of a triangle. We state the version with points on the sides, deliberately restricting to the acute case. There is also an “external” form of Ceva’s theorem that covers an intersection point outside the triangle.

Ceva’s theorem.

Let \(D\), \(E\), and \(F\) lie on \(BC\), \(CA\), and \(AB\) of triangle \(ABC\), respectively. The lines \(AD\), \(BE\), and \(CF\) are concurrent if and only if \[ AE\cdot BF\cdot CD = AF\cdot BD\cdot CE. \]

For altitudes, the right triangles \(ABE\) and \(ACF\) are similar at the acute angle, hence \[ \frac{AE}{AF} = \frac{AB}{AC}.\] Likewise, \[ \frac{BF}{BD} = \frac{BC}{BA}, \quad \frac{CE}{CD} = \frac{CB}{CA}. \] Multiplying these equalities gives Ceva’s condition.

Eighth proof — trigonometric Ceva

Besides the usual Ceva theorem one sometimes uses its trigonometric form, which is even slightly handier for altitudes. Again we state the interior case; with care the statement extends to the “external” configuration.

Trigonometric Ceva’s theorem.

Let \(D\), \(E\), and \(F\) lie on \(BC\), \(CA\), and \(AB\) of triangle \(ABC\), respectively. The lines \(AD\), \(BE\), and \(CF\) are concurrent if and only if \[ \frac{\sin \angle BAD}{\sin \angle CAD} \cdot \frac{\sin \angle CBE}{\sin \angle ABE} \cdot \frac{\sin \angle ACF}{\sin \angle BCF} = 1. \]

For altitudes the claim becomes obvious from the angle identities \[\angle BAD = \angle BCF = 90^\circ - \angle B,\] \[\angle CBE = \angle CAD = 90^\circ - \angle C,\] \[\angle ACF = \angle ABE = 90^\circ - \angle A.\]

Ninth proof — the dot product

We now turn to proofs that may look computational at first glance but are in fact quite symmetric and elegant. The first is the vector approach: perpendicularity means zero dot product. The key is the following beautiful identity.

Four-point formula.

For any four points \(A\), \(B\), \(C\), and \(D\) (in the plane or even in space), \[ \overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{AC} \cdot \overrightarrow{DB} + \overrightarrow{AD} \cdot \overrightarrow{BC} = 0. \]

The proof is standard: express every vector from \(A\),

\[ \overrightarrow{CD} = \overrightarrow{AD} - \overrightarrow{AC}, \quad \overrightarrow{DB} = \overrightarrow{AB} - \overrightarrow{AD}, \quad \overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}, \] substitute into the desired identity, and watch all terms cancel.

To deduce the orthocenter statement, take \(D\) to be the intersection of the altitudes from \(B\) and \(C\). Then the first two terms vanish, so the third is zero as well, meaning the altitude from \(A\) passes through the same point.

Incidentally, in space this yields: if two pairs of opposite edges of a tetrahedron are perpendicular, then so is the third pair. Such tetrahedra are called orthocentric because their four altitudes are concurrent.

Tenth proof — complex numbers

Complex numbers may not seem the natural tool here, yet the argument has its own symmetry. We identify points in the plane with complex numbers and denote them by the same letters.

Without loss of generality, place the vertices on the unit circle centered at \(0\), i.e. \(|A|=|B|=|C|=1\). This normalization is often helpful in purely complex solutions.

Orthocenter formula.

The orthocenter is given by \[ H = A + B + C. \]

This is a particularly pretty formula. To prove it, check for instance that \(AH\) is perpendicular to \(BC\). Equivalently, the complex number \(\frac{A-H}{B-C}\) should be purely imaginary, i.e. change sign under complex conjugation. First, using the formula for \(H\),

\[ z = \frac{A - H}{B - C} = \frac{-B - C}{B - C}, \]

and second,

\[ \overline{z} = \frac{-\overline{B} - \overline{C}}{\overline{B} - \overline{C}} = \frac{-1/B - 1/C}{1/B - 1/C} = \frac{-C - B}{C - B} = -z. \]

In the last step we used that for a point on the unit circle centered at \(0\), complex conjugation equals taking the reciprocal.

From this proof one easily gets the Euler line: the centroid has coordinate \((A+B+C)/3\) and lies on the segment joining \(0\) to the orthocenter, dividing it in the ratio \(1:2\).

Eleventh proof — Cartesian coordinates

Proving the claim in Cartesian coordinates may sound odd, but the argument below shares with the complex-number proof both a certain elegance and a useful corollary about the orthocenter that is not so easy to spot by other means. A brute-force coordinate proof is possible but tedious. Instead we give a more conceptual argument that assumes a bit more background.

The first idea is a clever choice of coordinates. As with complex numbers, the presentation benefits from a good setup. One might align axes with a side and the altitude to it; that works and is not too hard to finish. Here we impose a different condition: the vertices lie on the hyperbola \(xy=1\). The natural question is why this is legitimate — much like asking, in the previous proof, why we may assume the circumradius is \(1\). In fact, through the vertices of any triangle one can draw a rectangular hyperbola, and one can even fix the directions of its asymptotes.

Thus let the vertices \(A\), \(B\), and \(C\) have coordinates

\[ \left(x_a, \frac{1}{x_a}\right), \quad \left(x_b, \frac{1}{x_b}\right), \quad \left(x_c, \frac{1}{x_c}\right). \]

We claim that the point \(H\) with coordinates

\[ \left(-\frac{1}{x_a x_b x_c}, -x_a x_b x_c\right) \]

is the orthocenter. It suffices, say, to check that \(AH\) and \(BC\) are perpendicular, i.e. that the product of their slopes equals \(-1\). This reduces to the identity

\[ \frac{\frac{1}{x_a} + x_a x_b x_c}{x_a + \frac{1}{x_a x_b x_c}} \cdot \frac{\frac{1}{x_b} - \frac{1}{x_c}}{x_b - x_c} = -1, \]

which is verified by straightforward algebra.

What I like about this proof is, first, the neat coordinate formulas for the orthocenter, and second, the corollary we obtain along the way:

Corollary.

If a rectangular hyperbola passes through the vertices of a triangle, it also passes through its orthocenter.

Twelfth proof — projections

One of the strongest computational ways to prove parallelism or perpendicularity is to compare projections. Suppose the altitudes \(BE\) and \(CF\) of triangle \(ABC\) meet at \(H\). We show that \(AH\) is perpendicular to \(BC\), equivalently parallel to the altitude \(AD\). It is enough to check that the projections of \(AD\) and \(AH\) onto \(AB\) and \(AC\) have the same ratio (for directed segments this is the clean formulation). The projections of \(AH\) onto \(AB\) and \(AC\) are in the ratio \(AF/AE\), which by the obvious similarity of triangles \(ABE\) and \(ACF\) equals \(AC/AB\).

It remains to verify that the projections of \(AD\) onto those sides are inversely proportional to the side lengths. But they are inversely proportional to the sines of the corresponding angles

Thirteenth proof — isogonal conjugation

Finally, the last proof uses isogonal conjugation and is close in spirit to trigonometric Ceva. To prove that three rays through the vertices are concurrent, it suffices to prove the same for the rays symmetric to them in the angle bisectors at those vertices. For altitudes, these symmetric rays (the isogonal lines) are the rays toward the circumcenter, which are concurrent.

14, 15, 16,

The list can be continued. The more geometric tools you know, the wider your toolbox. Try to prove the altitude concurrency on your own using inversion, polars, or theorems on isogonal lines