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# 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 $\Gamma$ 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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