# Collinearity with the OrthocenterWhat Is This About? A Mathematical Droodle

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Explanation The applet below provides an illustration to a problem from an outstanding collection by T. Andreescu and R. Gelca (also from an earlier Russian book by I. Sharygin):

Let ABC be an acute triangle. The Points M and N are taken on the sides AB and AC, respectively. The circles with diameters BN and CM intersect at points P and Q. Prove that P, Q, and the orthocenter H are collinear. ### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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A clue to the solution lies in the observation that the feet of the altitudes from two vertices and these vertices themselves are concyclic; the circumcircle of the four points has the side joining the two vertices as a diameter.

So, let BHB and CHc be the altitudes to AC and AB, respectively. The orthocenter H is the intersection of the altitudes BHB and CHc. By the Intersecting Chords Theorem

BH × HHb = CH × HHc.

We have two circles, say, C(BN) and C(CM), with diameters BN and CM respectively. Assume line PH intersects C(BN) in Qb, and C(CM) in Qc. We want to show that Qb = Qc.

But in C(BN),

BH × HHb = PH × HQb

while in C(CM),

CH × HHc = PH × HQc.

By transitivity of equality, it follows that

PH × HQb = PH × HQc,

and subsequently, HQb = HQc, implying Qb = Qc, so that both coincide with Q.

Note that, as the solution shows, the requirement that ΔABC be acute is entirely spurious. Note also that the statement just proved is equivalent to the assertion that the radical axis of the circles on two cevians as diameters passes through the orthocenter of the triangle, making the orthocenter the radical center of three such circles.

### References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.3.8 (p. 12)
2. I. F. Sharygin, Problems in Geometry. Plane geometry, Nauka, Moscow, 1982, p. 46, problem 188 ### Power of a Point wrt a Circle 