A circle passes through the vertex of a rectangle $ABCD$ and touches its sides $AB$ and $AD$ at $M$ and $N$ respectively. If the distance from $C$ to the line segment $MN$ is equal to $5$ units, find the area of rectangle $ABCD$.
1. **Identify the vertex and setup the problem:**
   - Given a rectangle \(ABCD\) with a circle passing through vertex \(C\) and touching sides \(AB\) and \(AD\) at points \(M\) and \(N\) respectively.
   - The distance from \(C\) to the line segment \(MN\) is given as 5 units.

2. **Apply Theorem 1:**
   - Theorem 1 states: Let \(C, M, N\) be three points on a circle \(k\), and \(O\) the center of this circle \(k\). The tangents to the circle \(k\) at the points \(M\) and \(N\) meet at a point \(A\), and we denote by \(B, D\) and \(H\) the orthogonal projections of the point \(C\) on the lines \(AM, AN\) and \(MN\). Then, \(CB \cdot CD = CH^2\).

3. **Proof of Theorem 1:**
   - Since \(\angle CHM = 90^\circ\) and \(\angle CBM = 90^\circ\), the points \(H\) and \(B\) lie on the circle with diameter \(CM\). Thus, \(\angle CBH = \angle CMH\).
   - Since the line \(AN\) is the tangent to the circle \(k\) at the point \(N\), by the tangent-chordal angle theorem, \(\angle (CN; AD) = \angle CMN\). Thus, \(\angle CBH = \angle CMH = \angle CMN = \angle (CN; AD) = \angle CND\).
   - Since \(\angle CHN = 90^\circ\) and \(\angle CDN = 90^\circ\), the points \(H\) and \(D\) lie on the circle with diameter \(CN\), so that \(\angle CND = \angle CHD\). Thus, \(\angle CBH = \angle CHD\). Similarly, \(\angle CDH = \angle CHB\).
   - From \(\angle CBH = \angle CHD\) and \(\angle CHB = \angle CDH\), it follows that the triangles \(CBH\) and \(CHD\) are directly similar; thus, \(CB : CH = CH : CD\), so that \(CB \cdot CD = CH^2\).

4. **Apply Theorem 1 to the problem:**
   - In the given problem, points \(B\) and \(D\) are the orthogonal projections of point \(C\) on the lines \(AM\) and \(AN\) respectively.
   - Given \(CH = 5\) (the distance from \(C\) to the line segment \(MN\)), Theorem 1 yields:
     \[
     CB \cdot CD = CH^2 = 5^2 = 25
     \]
   - The area of the rectangle \(ABCD\) is given by \(CB \cdot CD\), which equals 25.

The final answer is \(\boxed{25}\)