Orthocenter on perpendicular bisector | INMO 2013

This is a problem from Indian National Mathematics Olympiad, INMO, 2013 based on Orthocenter on perpendicular bisector. Try out this problem.

Problem: Orthocenter on perpendicular bisector

In an acute angled triangle ABC with AB < AC the circle $latex \Gamma $ touches AB at B and passes through C intersecting AC again at D. Prove that the orthocenter of triangle ABD lies on $latex \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 $latex \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 $latex \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 $latex \angle BFD = \angle BED = 90^0 $ ). Hence $latex \angle FBH = \angle EDH $ (angle in the same segment FE). .... (ii)

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

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

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

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

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