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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 touches AB at B and passes through C intersecting AC again at D. Prove that the orthocenter of triangle ABD lies on 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 . 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 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 ). Hence (angle in the same segment FE). .... (ii)

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

Combining (ii) and (iii) we have implying HB = HC.

**Conversely** if we have HB = HC, this implies . Also (angles in the alternate segment subtended by chord BD)

Now consider triangles BEC and BFD. We have and . Therefore remaining angles BDF and EBC are also equal. But implying . Thus HDCB is cyclic. Hence proved.

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