Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National Olympiad Prove it Ireland

Let $H$ denote the orthocentre of $\triangle ABC$ , with circumcircle $\Gamma$ . The altitudes $AH$ , $BH$ , $CH$ intersect $\Gamma$ for the second time at the points $A'$ , $B'$ and $C'$ , respectively. The circle of diameter $AH$ intersects $\Gamma$ at the points $A$ and $D$ . Lines $DC$ and $A'B'$ intersect at $E$ , while lines $DB$ and $A'C'$ intersect at $F$ . Prove that $EF$ is parallel to $BC$ .

Solution

The idea of the proof is to show that EF is tangent at H to the circle which has AH as diameter. This will imply the result since AH is perpendicular to BC.

$\angle HDE = \angle MDC = \angle MAC = 90^\circ - \angle AMC = 90^\circ - \angle B.$

Because $AA'$ is perpendicular to $BC$ we also have
$90^\circ - \angle B = \angle BAA' = \angle BB'A' = \angle HB'E$
so that we have established $\angle HDE = \angle HB'E$ which implies that $DHEB'$ is cyclic. Then, using that $AA'$ is perpendicular to $BC$ and $BB'$ is perpendicular to $AC$ , we see
$\angle B'HE = \angle B'DE = \angle B'BC = 90^\circ - \angle C = \angle CAH.$
Finally, the intersection point L of the lines $BB'$ and $AC$ is on the circle with diameter AH since $BB' \perp AC$ . Thus the equality
$\angle LHE = \angle B'HE = \angle CAH = \angle LAH$
proves that EH is tangent to the circle with diameter AH. A similar proof shows that FH is also tangent to the circle with diameter AH, hence E, H, F are collinear and EF is tangent at H to the circle which has AH as diameter.

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