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Orthocenter on perpendicular bisector

INMO 2013

In an acute angled triangle ABC with AB < AC the circle \(\Gamma \) touches AB at B and passes through C intersecting AC again at D. Prove that the orthocenter of triangle ABD lies on \(\Gamma \) if and only if it lies on the perpendicular bisector of BC.

Discussion

RMO 2013 Q3Suppose H is the orthocenter of triangle ABD and it lies on the circle \(\Gamma \). We show that HB = HC (if we can show this then the perpendicular from H on BC will bisect BC).

DF and BE are altitudes of triangle ABD.

First we note that \(\angle FBH = \angle HCB \) for FB is tangent to the circle and angle made by a chord with a tangent is equivalent to an angle in the alternate segment. In this case the chord is BH.

Again FBDE is cyclic (since \(\angle BFD = \angle BED = 90^0 \) ). Hence \(\angle FBH = \angle EDH \) (angle in the same segment FE). …. (ii)

But HDCB is also cyclic (all vertices are on the circle). Hence \(\angle EDH = \angle HBC \) (exterior angle is equal to the interior opposite angle in a cyclic quadrilateral). …. (iii)

Combining (ii) and (iii) we have \(\angle HCB = HBC\) implying HB = HC.

Conversely if we have HB = HC, this implies \(\angle HBC = \angle HCB \) . Also \(\angle FBD = \angle DCB \) (angles in the alternate segment subtended by chord BD)

Now consider triangles BEC and BFD. We have \(\angle BEC = \angle BFD = 90^0 \) and \(\angle ECB = \angle FBD \). Therefore remaining angles BDF and EBC are also equal. But \(\angle DBC = \angle HCB \) implying \(\angle BDF = \angle HCB \). Thus HDCB is cyclic. Hence proved.

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December 2, 2013

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