Right triangle $ABC$ is inscribed in circle $W$. $\angle{CAB}=65$ degrees, and $\angle{CBA}=25$ degrees. The median from $C$ to $AB$ intersects $W$ and line $D$. Line $l_1$ is drawn tangent to $W$ at $A$. Line $l_2$ is drawn tangent to $W$ at $D$. The lines $l_1$ and $l_2$ intersect at $P$
Determine $\angle{APD}$
1. **Identify the given information and the goal:**
   - Right triangle \(ABC\) is inscribed in circle \(W\).
   - \(\angle CAB = 65^\circ\) and \(\angle CBA = 25^\circ\).
   - The median from \(C\) to \(AB\) intersects \(W\) at point \(D\).
   - Lines \(l_1\) and \(l_2\) are tangents to \(W\) at points \(A\) and \(D\) respectively, and they intersect at point \(P\).
   - We need to determine \(\angle APD\).

2. **Determine the center \(O\) of circle \(W\):**
   - Since \(ABC\) is a right triangle inscribed in circle \(W\), the hypotenuse \(AB\) is the diameter of the circle. Therefore, \(O\) is the midpoint of \(AB\).

3. **Analyze the cyclic quadrilateral \(AODP\):**
   - Since \(O\) is the center of the circle and \(D\) lies on the circle, \(AODP\) forms a cyclic quadrilateral.
   - In a cyclic quadrilateral, the opposite angles sum to \(180^\circ\).

4. **Calculate \(\angle AOD\):**
   - Since \(O\) is the center, \(\angle AOD\) is a central angle.
   - The measure of the arc \(BC\) is twice the measure of \(\angle CAB\) because \(\angle CAB\) subtends arc \(BC\).
   - Therefore, the measure of arc \(BC = 2 \times 65^\circ = 130^\circ\).
   - The central angle \(\angle BOC\) is equal to the measure of arc \(BC\), so \(\angle BOC = 130^\circ\).

5. **Determine \(\angle APD\):**
   - Since \(AODP\) is cyclic, \(\angle APD = 180^\circ - \angle AOD\).
   - \(\angle AOD\) is the same as \(\angle BOC\) because both are central angles subtending the same arc \(BC\).
   - Therefore, \(\angle AOD = 130^\circ\).
   - Thus, \(\angle APD = 180^\circ - 130^\circ = 50^\circ\).

The final answer is \(\boxed{50^\circ}\).