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*

Suppose 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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