Sangaku: Show line segment is perpendicular to diameter of container circle "From a 1803 Sangaku found in Gumma Prefecture. The base of an isosceles triangle sits on a diameter of the large circle. This diameter also bisects the circle on the left, which is inscribed so that it just touches the inside of the container circle and one vertex of the triangle. The top circle is inscribed so that it touches the outsides of both the left circle and the triangle, as well as the inside of the container circle. A line segment connects the center of the top circle and the intersection point between the left circle and the triangle. Show that this line segment is perpendicular to the drawn diameter of the container circle. (T. Rothman)"

Source: http://hermay.org/jconstant/wasan/sangaku/index.html
Enjoy!
EDIT
Note 1: The radius of the smaller circle touching the left vertex of the triangle is not necessarily half the radius of the enclosing circle. There IS variability in the base of the triangle. 
Note 2: Two of the vertices of triangle necessarily are on the enclosing circle.
 A: Let $A$ be the big circle with centre $O$ and diameter $PQ$
Let $B$ be the circle internally tangent to $A$ at $P$ and intersecting $PQ$ again at $M$
Let $R$ be on $A$ such that $\overline{MR} = \overline{QR}$
Let $C$ be the circle that is tangent to $A$, $B$, $MR$ and inside $A$ and outside both $B$ and completely on the same side of $PQ$ as $R$
Invert at $M$, mapping $P$ to $Q$
Then $A$ maps to $A$, $B$ maps to the line $T$ that is tangent to $B$ at $Q$, $MR$ maps to $MR$, and thus $C$ maps to the circle $D$ that is tangent to $A$, $T$, $MR$ and outside $A$ and on the same side of $T$ as $M$ and completely on the other side of $PQ$ as $R$
Let $K$ be the point on $MQ$ such that $RK \perp MQ$
Let $N$ be the point such that $PQ \perp MN$ and $\overline{MN} = 2\overline{MR}$ and $N$, $R$ are on the opposite side of $PQ$
Let $X$ be the intersection of $NO$ and $A$ between $N$ and $O$
Let $Y$ be the point on $T$ such that $NY \perp T$
Let $Z$ be the point on $MR$ such that $NZ \perp MR$
Let $\overrightarrow{MO} = r \overrightarrow{OQ}$ and WLOG $\overline{OQ} = 1$
Then $\overline{MR}^2 = \overline{RK}^2 + \overline{MK}^2 = \overline{OR}^2 - \overline{OK}^2 + \overline{MK}^2 = 1 - (\frac{1-r}{2})^2 + (\frac{1+r}{2})^2 = 1+r$
Thus $\overline{NO}^2 = \overline{MN}^2 + r^2 = 4 \overline{MR}^2 + r^2 = 4+4r+r^2 = (2+r)^2$
Thus $\overline{NX} = (2+r)-1 = \overline{MQ}$
Also it is clear that $\overline{NY} = \overline{MQ}$
Also since $\triangle MNZ \sim \triangle RMK$, $\overline{NZ} = 2 \overline{MK} = \overline{MQ}$
Since $D$ is uniquely defined and $N$ is the centre of an identically defined circle, $N$ is the centre of $D$
Since the centres of $C$ and $D$ are collinear with $M$, the centre of $C$ lies on $MN$, therefore the line through both $M$ and the centre of $C$ is perpendicular to $PQ$
(QED)
Note that the same solution applies to both cases where $C$ is on either side of $PQ$, with very minor changes. Inversion seems to simplify most of these kind of questions. Quite a few can be found online.
A: For those who do not like inversion... =P
Let $A$ be the big circle with centre $O$ and diameter $PQ$
Let $B$ be the circle internally tangent to $A$ at $P$ and intersecting $PQ$ again at $M$ and having centre $J$
Let $R$ be on $A$ such that $\overline{MR} = \overline{QR}$
Let $C$ be the circle that is tangent to $A$, $B$, $MR$ and inside $A$ and outside both $B$ and completely on the same side of $PQ$ as $R$
Let $K$ be the point on $MQ$ such that $RK \perp MQ$
Let $D$ be the circle tangent to $B$ and $MR$ with centre $N$ such that $PQ \perp MN$ and $N$,$R$ are on the same side of $PQ$
Let $Z$ be the point where $D$ is tangent to $MR$
Let $\overrightarrow{MO} = r \overrightarrow{OQ}$ and WLOG $\overline{OQ} = 1$
Let $x = \overline{MN}$
Let $y$ be the radius of $D$
Then $\overline{MR}^2 = \overline{RK}^2 + \overline{MK}^2 = \overline{OR}^2 - \overline{OK}^2 + \overline{MK}^2 = 1 - (\frac{1-r}{2})^2 + (\frac{1+r}{2})^2 = 1+r$
Since $\triangle MNZ \sim \triangle RMK$, $\frac{x}{y} = \overline{MN} / \overline{NZ} = \overline{RM} / \overline{MK} = \sqrt{1+r} / \frac{1+r}{2} = \frac{2}{\sqrt{1+r}}$
Thus $(1+r)x^2 = 4y^2$
Also $(y+\frac{1-r}{2})^2 = \overline{NJ}^2 = \overline{MN}^2 + \overline{MJ}^2 = (\frac{1-r}{2})^2 + x^2$
Thus $(1+r)y^2 + (1+r)(1-r)y = (1+r)x^2 = 4y^2$
Since $y\not=0$, $(3-r)y = 1-r^2$
$\overline{OQ} = y + \overline{ON}$
$\Leftrightarrow (1-y)^2 = \overline{ON}^2 = \overline{MN}^2 + \overline{MO}^2 = x^2 + r^2$
$\Leftrightarrow (1+r)(1-2y+y^2) = 4y^2 + (1+r)r^2$
$\Leftrightarrow (3-r)y^2 + (2+2r)y - (1+r)(1-r^2) = 0$
$\Leftrightarrow (1-r^2)y + (2+2r)y = (1+r)(1-r^2)$
$\Leftrightarrow (1+r)(3-r)y = (1+r)(1-r^2)$ which is clearly true
Thus $D$ is tangent to $A$
Since $C$ is uniquely defined and $D$ is an identically defined circle, $N$ is the centre of $C$
Therefore the line through both $M$ and the centre of $C$ is perpendicular to $PQ$
(QED)
