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Classical

Four Orthocenters of a Cyclic Quadrilateral

A hidden half-turn

For a cyclic quadrilateral, the four orthocenters obtained by omitting one vertex at a time are the images of the original vertices under a single half-turn.

Methods Vectors · Central symmetry

The theorem

Theorem

Let ABCDABCD be a cyclic quadrilateral. Let P,Q,R,SP,Q,R,S be the orthocenters of the triangles BCD,CDA,DAB,ABCBCD,CDA,DAB,ABC, respectively.

Then the segments

AP,BQ,CR,DSAP,\qquad BQ,\qquad CR,\qquad DS

have a common midpoint MM. Equivalently, the half-turn centered at MM sends

A↦P,B↦Q,C↦R,D↦S.A\mapsto P,\qquad B\mapsto Q,\qquad C\mapsto R,\qquad D\mapsto S.

In particular,

ABCD≅PQRS.ABCD\cong PQRS.
A cyclic quadrilateral ABCD and the quadrilateral PQRS formed by the orthocenters of BCD, CDA, DAB, and ABC; the four segments AP, BQ, CR, and DS share the midpoint M.
Figure 1. The half-turn centered at M sends A, B, C, D to P, Q, R, S, respectively.

Proof

Let OO be the center of the circumcircle of ABCDABCD, and take OO as the origin. Write

a,b,c,d\mathbf a,\qquad \mathbf b,\qquad \mathbf c,\qquad \mathbf d

for the position vectors of A,B,C,DA,B,C,D.

For a triangle inscribed in a circle centered at the origin, the orthocenter formula says that its position vector is the sum of its three vertex vectors. Therefore,

p=b+c+d,q=c+d+a,r=d+a+b,s=a+b+c.\begin{aligned} \mathbf p&=\mathbf b+\mathbf c+\mathbf d,\\ \mathbf q&=\mathbf c+\mathbf d+\mathbf a,\\ \mathbf r&=\mathbf d+\mathbf a+\mathbf b,\\ \mathbf s&=\mathbf a+\mathbf b+\mathbf c. \end{aligned}

Set

m=a+b+c+d2.\mathbf m = \frac{\mathbf a+\mathbf b+\mathbf c+\mathbf d}{2}.

Then

p=2m−a,q=2m−b,r=2m−c,s=2m−d.\begin{aligned} \mathbf p&=2\mathbf m-\mathbf a,\\ \mathbf q&=2\mathbf m-\mathbf b,\\ \mathbf r&=2\mathbf m-\mathbf c,\\ \mathbf s&=2\mathbf m-\mathbf d. \end{aligned}

Thus MM is the midpoint of each of AP,BQ,CR,DSAP,BQ,CR,DS. The half-turn, or central symmetry, centered at MM therefore sends A,B,C,DA,B,C,D to P,Q,R,SP,Q,R,S, respectively. In particular, the two quadrilaterals are congruent.

Source

This configuration appeared as Balkan Mathematical Olympiad 1984 (BMO 1984), Problem 2, where the stated conclusion was the congruence of the two quadrilaterals. The formulation above records the stronger half-turn relation given by the proof.