Problem 1531

1. Let $\ell$ be the angle bisector of $\angle ADC$. Since $D$ is the midpoint of $BC$, we have $BD = DC$.
2. Let $X \in \ell$ such that $AX > BX$. Also, let $\ell \cap AC = Y$.
3. Since $\ell$ is the angle bisector of $\angle ADC$, we have $\angle ADY = \angle YDC$.
4. Given that the angle bisector $\ell$ is tangent to the circumcircle of triangle $ABD$ at $D$, we know that $\angle ADY = \angle XDB$ (by the tangent-secant angle theorem).
5. Since $\angle XDB = \angle DAB$ (as $D$ is the point of tangency), we have:
$$\angle ADY = \angle DAB$$
6. From the above, we can infer that $\angle ABC = \angle ADY$.
7. Since $D$ is the midpoint of $BC$, we have $BD = DC$. Also, since $\angle ADY = \angle DAB$, it implies that $\triangle ABD$ is isosceles with $AD = BD$.
8. Therefore, $BD = DC = AD$, making $\triangle ABD$ and $\triangle ADC$ isosceles with $AD$ as the common side.
9. Since $\angle ADY = \angle DAB$ and $\angle ABC = \angle ADY$, we have:
$$\angle ABC = \angle DAB$$
10. Given that $\angle ABC = \angle DAB$ and $BD = DC = AD$, it follows that $\angle A = 90^\circ$.

$\blacksquare$