$\left.\begin{array}{l}{[\quad \text { Angle between the tangent and the chord }} \\ {[\text { Pythagorean Theorem (direct and inverse) }}\end{array}\right]$

In quadrilateral $A B C D$, diagonals $A C$ and $B D$ are perpendicular and intersect at point $P$. The length of the segment connecting vertex $C$ to point $M$, which is the midpoint of segment $A D$, is $\frac{5}{4}$. The distance from point $P$ to segment $B C$ is $\frac{1}{2}$ and $A P=1$. Find $A D$, given that a circle can be circumscribed around quadrilateral $A B C D$.
Let the line $M P$ intersect the segment $B C$ at point $K$. Denote $\angle A D B = \angle A C B = \alpha$. Since $P M$ is the median of the right triangle $A P D$ drawn from the vertex of the right angle, then

$$
P M = M A = M D, \angle B P K = \angle D P M = \angle A D B = \alpha,
$$

and since $\angle C B P = 90^{\circ} - \alpha$, then

$$
\angle B K P = 180^{\circ} - \alpha - (90^{\circ} - \alpha) = 90^{\circ}
$$

i.e., $P K \perp B C$. Therefore, $P K = \frac{1}{2}$.

From the right triangles $A P D$ and $C K P$, we find that

$$
M P = \frac{1}{2} A D = \frac{1}{2} \cdot \frac{A P}{\sin \alpha^{\prime}} = \frac{1}{2 \sin \alpha^{\prime}}, C K = K P \operatorname{ctg} \alpha = \frac{1}{2} \operatorname{ctg} \alpha,
$$

so

$$
M K = M P + K P = \frac{1}{2 \sin \alpha} + \frac{1}{2}.
$$

Applying the Pythagorean theorem to the right triangle $M K C$, we get the equation

$$
\left(\frac{1}{2 \sin \alpha} + \frac{1}{2}\right)^2 + \left(\frac{1}{2} \operatorname{ctg} \alpha\right)^2 = \frac{25}{16}
$$

from which we find that $\frac{1}{\sin \alpha} = \frac{3 \sqrt{6} - 2}{t}$. Therefore,

$$
A D = \frac{A P}{\sin \alpha^{\prime}} = \frac{1}{\sin \alpha^{\prime}} = \frac{3 \sqrt{6} - 2}{t}
$$