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Two Views of the Orthocenter

Cyclic quadrilaterals, vectors, and the Euler line

Two cyclic quadrilaterals explain why the third altitude is forced through the intersection of the first two, while a circumcenter-based vector formula reveals the Euler line.

Methods Cyclic quadrilaterals · Vectors

The theorem

Theorem

Existence of the orthocenter

The three altitudes of a triangle ABCABC are concurrent. Their common point is called the orthocenter of ABCABC.

An acute triangle ABC with D, E, and F the perpendicular feet from A, B, and C, respectively. The altitudes AD, BE, and CF meet at H.
The three altitudes of triangle ABC meet at H.

First view: two cyclic quadrilaterals

Let EE and FF be the feet of the altitudes from BB and CC on the lines ACAC and ABAB, respectively:

BE⊥AC,CF⊥AB.BE \perp AC, \qquad CF \perp AB.

Let

H=BE∩CF,D=AH∩BC.H=BE\cap CF,\qquad D=AH\cap BC.

Because

∠BFC=∠BEC=90∘,\angle BFC=\angle BEC=90^\circ,

the four points F,E,C,BF,E,C,B lie on a circle. Also,

∠AFH=∠AEH=90∘,\angle AFH=\angle AEH=90^\circ,
An acute triangle ABC with D, E, and F the altitude feet from A, B, and C, respectively. The altitudes AD, BE, and CF meet at H, and the circles through B, C, E, F and through A, E, F, H are shown.
Figure 1. The circles ω₁ = (BCEF) and ω₂ = (AEFH) carry the angle information that forces the third altitude.

so the four points A,F,H,EA,F,H,E are also concyclic. In the configuration shown, the rays CDCD and CBCB, EBEB and EHEH, and AHAH and ADAD coincide. The two circles therefore give

∠FCD=∠FCB=∠FEB=∠FEH=∠FAH=∠FAD.\angle FCD =\angle FCB =\angle FEB =\angle FEH =\angle FAH =\angle FAD.

By the converse of the inscribed-angle theorem, F,A,C,DF,A,C,D are concyclic. Hence

∠ADC=∠AFC=90∘.\angle ADC=\angle AFC=90^\circ.

Therefore AD⊥BCAD\perp BC. The point HH lies on all three altitudes, and hence is the orthocenter of ABCABC.

Second view: the circumcenter as origin

The vector proof begins from a different point of view. Instead of intersecting two altitudes and verifying the third, it constructs a point that lies on all three altitudes at once.

Let OO be the circumcenter of ABCABC, and take OO as the origin. Write

OA→=a,OB→=b,OC→=c.\overrightarrow{OA}=\mathbf a,\qquad \overrightarrow{OB}=\mathbf b,\qquad \overrightarrow{OC}=\mathbf c.

Because A,B,CA,B,C lie on the circle centered at OO,

∥a∥=∥b∥=∥c∥.\lVert\mathbf a\rVert=\lVert\mathbf b\rVert=\lVert\mathbf c\rVert.

Define a point HH by

OH→=h=a+b+c.\overrightarrow{OH}=\mathbf h=\mathbf a+\mathbf b+\mathbf c.

We first show that AH⊥BCAH\perp BC. We have

AH→=h−a=b+c,BC→=c−b.\overrightarrow{AH}=\mathbf h-\mathbf a=\mathbf b+\mathbf c, \qquad \overrightarrow{BC}=\mathbf c-\mathbf b.

Therefore

AH→⋅BC→=(b+c)⋅(c−b)=∥c∥2−∥b∥2=0.\begin{aligned} \overrightarrow{AH}\mathbin{\cdot}\overrightarrow{BC} &=(\mathbf b+\mathbf c)\mathbin{\cdot}(\mathbf c-\mathbf b)\\ &=\lVert\mathbf c\rVert^2-\lVert\mathbf b\rVert^2\\ &=0. \end{aligned}

Hence AH⊥BCAH\perp BC. The same calculation, applied cyclically, gives

BH⊥CA,CH⊥AB.BH\perp CA,\qquad CH\perp AB.

Thus HH is the orthocenter of ABCABC.

The choice of origin is the key: because the vertex vectors have equal lengths, a sum such as b+c\mathbf b+\mathbf c is perpendicular to the corresponding difference c−b\mathbf c-\mathbf b. This makes the symmetric formula for h\mathbf h natural.

A scalene triangle ABC on its circumcircle with O as origin. The equal-length vertex vectors a, b, c are drawn from O, the orthocenter has vector h = a + b + c, and the centroid G lies on OH.
Figure 2. With the circumcenter as origin, the vertex vectors have equal length and the orthocenter is given by h = a + b + c. The centroid G lies on OH.

The Euler line

Let GG be the centroid of ABCABC. With the same origin OO, its position vector is

OG→=g=a+b+c3.\overrightarrow{OG}=\mathbf g=\frac{\mathbf a+\mathbf b+\mathbf c}{3}.

Comparing this with the vector of HH gives

g=13h.\mathbf g=\frac13\mathbf h.

Consequently, O,G,HO,G,H are collinear. Unless the triangle is equilateral,

OG:GH=1:2.OG:GH=1:2.

In the equilateral case, the circumcenter, centroid, and orthocenter coincide.

The vector proof therefore contains the Euler line without an additional construction.

The two proofs are complementary. The synthetic proof shows how the two cyclic quadrilaterals created by the right angles force the third altitude through the intersection of the first two; the vector proof constructs the orthocenter symmetrically and leads directly to the Euler line.

The Centroid through Ceva and Vectors develops Ceva, isotomic conjugation, and the explicit vector construction of the centroid.

The Euler Line from the Medial Triangle gives a synthetic homothety proof of the same collinearity obtained here by vectors.

Four Orthocenters of a Cyclic Quadrilateral applies the orthocenter vector formula four times and reveals a hidden half-turn.

The Nine-Point Circle from the Circumcircle uses the unit-circle formula for the orthocenter to recover all nine classical points.

Simson’s Theorem uses two cyclic quadrilaterals to align three perpendicular feet, then extends the same argument to oblique projections.

Crux Mathematicorum Problem 5098 uses an orthocenter in a more specialized cyclic configuration.

Diagram and construction source

The three figures and their complete tkz-euclide sources are available in the diagram and construction record.