In this figure, prove that $H$ lies on the circle iff it lies on the perpendicular bisector of $\Delta ABC$ 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. 

 A: Angle-chasing is helpful.
Let $Q$ be the orthocenter of $\triangle ABD$, the common point on altitudes $\overline{AA^\prime}$, $\overline{BB^\prime}$, $\overline{DD^\prime}$. Let $G$ be the point where $\overline{DD^\prime}$ meets the circle.

Right triangles $\triangle ABA^\prime$ and $\triangle DBD^\prime$ overlap at $\angle B$, so their "other" acute angles match, with, say, measure $\theta$; likewise, the non-overlapped acute angles of $\triangle ADA^\prime$ and $\triangle BDB^\prime$, with measure $\phi$. In the circle, inscribed angles $\angle GDB^\prime$ and $\angle GBD^\prime$ subtend the same arc, $\stackrel{\frown}{BG}$, so that $|\angle GBD^\prime| = |\angle GDB| = \theta$.
Now the implication chain can begin:
$$\begin{align}
Q \text{ lies on the circle } \quad &\Leftrightarrow \quad Q \text{ and } G \text{ coincide[*]} \\
&\Leftrightarrow \quad |\angle QBG| = 0 \\
&\Leftrightarrow \quad |\angle DBA| = |\angle DAB| = \theta + \phi \\
&\Leftrightarrow \quad \triangle ADB \text{ is isosceles with base } \overline{AB} \\
&\Leftrightarrow \quad \text{altitude } \overline{DD^\prime} \text{ is also a perp. bis. of } \overline{AB} \\
&\Leftrightarrow \quad Q \text{ lies on the perp. bis. of } \overline{AB}
\end{align}$$
[*] Why can't $Q$ and $D$ coincide?
A: Hint:

*

*You want to prove that $H$ lies on the bisector of $AC$.

*$A$ is the orthocenter of $\triangle BDH$.

*Prove the following lemma:

For any triangle $XYZ$, the three reflections $H_{XY}$, $H_{YZ}$, $H_{ZX}$ of an orthocenter $H$ (with respect to sides) are concyclic with $X$, $Y$ and $Z$.
  
*Try to reverse the following argument: $|\angle ABH| = |\angle CBH|$ because $|\angle ABH| = |\angle ADH|$ (because $\angle BHD$ is common to appropriate right triangles) and $|\angle CBH| = |\angle CDH|$ (assuming $BDCH$ is cyclic, depending on the position of $H$, some angles might be "flipped").
  
  I hope this helps $\ddot\smile$

