NZMO Round Two

Geometry Difficulty 6.6 National Olympiad Prove it New Zealand

Problem:

Let $ABC$ be an acute scalene triangle with $AC > BC > AB$ . Let the orthocentre be $H$ and circumcentre be $O$ . Suppose that lines $BO$ and $CH$ intersect at a point $D$ . Point $E$ (where $E \neq C$ ) lies on side $AC$ so that $OECD$ is cyclic. Point $F$ (where $F \neq C$ ) lies on side $BC$ such that $CE = FE$ . Prove that $B H D F$ is cyclic.

(The orthocentre of a triangle is the point of intersection of its altitudes.)

Solution

Solution:

Let $\alpha = \angle BAC$ . Let $BB'$ and $CC'$ be altitudes in triangle $ABC$ , as shown.

Claim. Triangle $CDE$ is isosceles with $CE = DE$ .

Proof.

$\angle DEC = \angle DOC \\ \qquad = \angle BOC \\ \qquad = 2\angle BAC \\ \qquad = 2\alpha$

$\angle DCE = \angle C'CA = 90^\circ - \angle C'AC \\ \qquad = 90^\circ - \alpha$

$\begin{array}{rl} & \angle CDE = 180^\circ - \angle DEC - \angle DCE \\ & \qquad = 180^\circ - 2\alpha - (90^\circ - \alpha) \\ & \qquad = 90^\circ - \alpha \\ & \qquad = \angle DCE. \end{array} \quad \text{(angle sum in } \triangle CDE)$

Since $CE = DE$ and $CE = FE$ , we have $CE = DE = FE$ , therefore $E$ is the circumcentre of triangle $CDF$ . Consequently,

$\begin{array}{rl} & \angle HDF = 180^\circ - \angle CDF \\ & \qquad = 180^\circ - \frac{1}{2} \cdot \angle CEF \\ & \qquad = 180^\circ - \frac{1}{2} \bullet (180^\circ - 2\angle ECF) \\ & \qquad = 180^\circ - \frac{1}{2} \bullet (180^\circ - 2\angle B'CB) \\ & \qquad = 180^\circ - (90^\circ - \angle B'CB) \\ & \qquad = 180^\circ - \angle B'BC \\ & \qquad = 180^\circ - \angle HBF \end{array} \quad \text{(angles in } \triangle B'BC)$

so $B H D F$ is cyclic.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.