Area of a circle passing through two vertices of a parallelogram touching one edge. For reference:

Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of ​​the circle. (Answer: $12\pi$)


My progress:
$AD^2 =AI\cdot AC \implies 10^2=AI\cdot14$
$ \implies AI = \dfrac{50}{7} ,\  IC = \dfrac{48}{7}$
In $\triangle ABC$
$14^2=6^2+10^2-2\cdot6\cdot10\cdot\cos \angle B\implies
\cos \angle B =-\dfrac{1}{2}\therefore \angle B =120^o  $
....I can't see many options????

Drop the perp from $A$ to $BC$ and call the foot $E$.

Let $BE=x$, then using Pythagoras' theorem, $$6^2-x^2=14^2-(10+x)^2\implies x=3.$$
Therefore in right triangle $ABE$, $\angle BAE=30^\circ$ and so is $\angle CDO$.
Let $M$ be the midpoint of side $CD$, then considering $\triangle ODM$, $OD$, or, radius of the circle is $2\sqrt3$.
Hence, the area of circle is $12\pi$.
