# Collinearity with the Orthocenter

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

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.

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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 BH_{B} and CH_{c} be the altitudes to AC and AB, respectively. The orthocenter H is the intersection of the altitudes BH_{B} and CH_{c}. By the Intersecting Chords Theorem

BH × HH_{b} = CH × HH_{c}.

We have two circles, say, C(BN) and C(CM), with diameters BN and CM respectively. Assume line PH intersects C(BN) in Q_{b}, and C(CM) in Q_{c}. We want to show that _{b} = Q_{c}.

But in C(BN),

BH × HH_{b} = PH × HQ_{b}

while in C(CM),

CH × HH_{c} = PH × HQ_{c}.

By transitivity of equality, it follows that

PH × HQ_{b} = PH × HQ_{c},

and subsequently, HQ_{b} = HQ_{c}, implying Q_{b} = Q_{c}, 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

- T. Andreescu, R. Gelca,
*Mathematical Olympiad Challenges*, Birkhäuser, 2004, 5^{th}printing, 1.3.8 (p. 12) - I. F. Sharygin,
*Problems in Geometry. Plane geometry*, Nauka, Moscow, 1982, p. 46, problem 188

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny