Try to solve it in a different way

In $\triangle{ABC}$ , $AC$ is the hypotenuse, so $\angle{ABC} = {90}^{\circ}$ .
The minimum number of points required for a circle to pass through = $3$ points.
Draw the circumcircle of $\triangle{ABC}$ .
Now, $\angle{ABC} = {90}^{\circ}$ .
The angle inside a semicircle is a right angle.
So, $AC$ is the diameter of the circle.
$\therefore$ $D$ is the center of the circle. (Mid-point of the diameter of the circle is the center of the circle)
$\therefore \angle{BDC} = 2\angle{BAC}$ (Angle at the centre is twice the angle at the remaining part of the circle).
$\implies \angle{BDC}$ = $2 × {30}^{\circ}\\ = \boxed{{60}^{\circ}}$ .

$\large \boxed{\text{Q.E.D}}$

Use sine rule in both small triangles as follows:- Given $AD=DC= x$ (say) Now, $\dfrac{x}{\sin (ABD)} = \dfrac{BD}{sin30º}\\ \implies BD =\dfrac{(x × sin30º)}{sin (ABD)}.\longrightarrow(1)$ Also, in the other triangle:- $\dfrac{x}{\sin(90-ABD)} = \dfrac{BD}{sin60º}\\ \implies BD = \dfrac{(x.sin60º)}{cos(ABD)} \longrightarrow (2)$ Equating (1) & (2).. you get:- $\tan (ABD)= \dfrac{1}{\sqrt 3}\\ \implies \angle ABD=30º$ Now $\angle BDC = \angle ABD + 30º$ (exterior angle property) = $\boxed{60º}$ . Thank you.