# 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

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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