In an acute-angled triangle $ABC$ with $AB < AC$, the circle $\Gamma$ touches $B$ and intersects $AC$ at $D$ and then passes through $C$. Prove that the orthocentre of $\Delta ABD$ lies $\Gamma$ if and only if it lies on the perpendicular bisector of $\Delta ABC$.
I tried to first prove that if $H$ lies on the perpendicular bisector, $\Gamma$. Assume that for some $\Delta ABC$, $H$ lies on the perpendicular bisector. Consider the following figure. The obvious thing to prove was that $HDCB$ is cyclic.
Now, let $\angle FBH = \alpha, \angle HBD = \alpha$. As $AB$ is tangent to $\Gamma$, by the special case of the inscribed angle theorem, $\angle FBD = \angle BCD = \alpha + \beta$
Construct through $B$ a line parallel to $HD$. Let it meet the circle at $K$ (not shown in fig). If we assume that $HDCB$ is cyclic, then $\angle DHB = 180 - \alpha - \beta$. But this, in turn would mean that $\angle HBK = \alpha + beta \implies \angle DBK = \beta$. But this would mean that $\angle HDB = \beta$.
Now, in $\Delta FBH$, it is clear that $\angle FHB = 90 - \beta = \angle GHD$. This means $\angle ADF = \beta$. Now this would mean that $DF$ is an altitude as well as an angle bisector, which is only possible if $\Delta ABD$ is isosceles. So, the problem reduces to:
Prove that if $H$ lies on the perpendicular bisector, $HI$, $\Delta ABD$ is isosceles.
I couldn't prove this.
I also started over and resorted to angle-chasing, but could get nowhere. I couldn't think of any way to use the fact that $BI =IC$, and so couldn't use all the data.