Consider the circle $O$ of diameter $AB$, and an infinite line that intersects the perpendicular $AB$ at the point $I$ between $O$ and $B$. Let $M$ be any point on the circle. Lines $AM$ and $BM$ intersect the perpendicular line at $C$ and $D$ respectively.
- Show that triangles $ACI$ and $DBI$ are similar.
- The circumscribed circle $ACD$ to the triangle $ACD$, intersects $AB$ at $B'$. Show that triangle $BDB'$ is isosceles. On which line does the center $S$ of the circle move when $M$ describes the circle $O$?
$\triangle DBI(\text{right}) \implies \angle D =90^\circ -\angle B=\angle A$
$ \triangle ACI: \angle ACI = 90^\circ-\angle A = \angle B \therefore \triangle ACI\sim \triangle DBI(S.S.)$
Item 2 could not be demonstrated, but by geogebra the circumscribed circle passes through the center of the other circle.
Part II
$\angle B = \angle C = 90^{\circ} - \angle A$ ${\space}$ (Sum of the $\angle$`s in $\triangle ACI$ and $\triangle AMB$ )
$\angle B' = \angle C =90^{\circ} - \angle A $ ${\space}$(exterior $\angle$ of cyclic quadrilateral ACDB')
$DB=DB'$ ${\space}$ (sides are $=$ opposite $= \angle$`s)
$\triangle BDB'$ is an isosceles $\triangle$
Claim : The centre of the cicle S is on the perpendicular bisector of $AB'$
Proof : Since point D on on the line $I$ that is perpendicular to line $AB$ and $\angle B=\angle B'$ then $BI=B'I$ and $B'$ is fixed on $AB$ for different $M$`s but same circle centered $O$.
Point $B'$ is on the circle passing through $ACD$ that is a chord of circle $ACD$.
Centre of the circle $S$ is moving on the perpendicular bisector of $AB'$