Question

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 orthocentre of triangle $ABD$ lies on the circle $\Gamma$ if and only if it lies on the perpendicular bisector of $BC.$

Answer 1

$\angle ADB=\angle B$
$\therefore △ADB$ and $△ABC$ are similar.
In particular, ABD is an acute-angled triangle.
Let H denote the orthocenter of triangle ABD.
$\therefore \angle BHD={180}^{\circ }-\angle A$ .
Suppose that H lies on $\Gamma$ .
Since AB < AC the point D lies on the segment AC and $\angle C={180}^{\circ }\angle BHD=\angle A$ .
$\therefore BH$ is the perpendicular bisector of $AC$.
Hence $\angle HBC=\angle ABC=\angle HCB$ , so H lies on the perpendicular bisector of BC.
Conversely, suppose that H lies on the perpendicular bisector of BC.
Then $\angle HCB=\angle HBC={90}^{\circ }\angle C$.
Since $\angle ABD=\angle C$ it follows that $\angle HDB={90}^{\circ }\angle C$ .
Since $\angle HCB=\angle HDB$ we hae that H lies on $\Gamma$ .