A convex pentagon $ ABCDE$ is inscribed in a circle. The distances of $ A$ from the lines $ BC,CD,DE$ are $ a,b,c,$ respectively. Compute the distance of $ A$ from the line $ BE$.
To solve the problem, we will use the lemma provided and apply it to the given pentagon \(ABCDE\) inscribed in a circle. The lemma states that for a cyclic quadrilateral \(ABCD\) with a point \(P\) on its circumcircle, the product of the distances from \(P\) to opposite sides is equal. We will extend this idea to the pentagon \(ABCDE\).

1. **Identify the relevant distances and angles:**
   - Let \(A\) be the point from which we measure the distances to the lines \(BC\), \(CD\), and \(DE\), denoted as \(a\), \(b\), and \(c\) respectively.
   - We need to find the distance from \(A\) to the line \(BE\).

2. **Apply the lemma to the cyclic quadrilateral \(ABCD\):**
   - Consider the quadrilateral \(ABCD\) with \(A\) on its circumcircle.
   - Let \(X\), \(Y\), \(Z\), and \(V\) be the orthogonal projections of \(A\) on the sides \(AD\), \(AB\), \(BC\), and \(CD\) respectively.
   - According to the lemma, \(AX \cdot AZ = AY \cdot AV\).

3. **Relate the distances using the lemma:**
   - Since \(A\) is on the circumcircle, the angles \(\angle PAB\) and \(\angle PCD\) are equal.
   - Similarly, \(\angle PAX\) and \(\angle PCD\) are equal.
   - Therefore, \(AX \cdot AZ = AY \cdot AV\).

4. **Extend the lemma to the pentagon \(ABCDE\):**
   - Now consider the pentagon \(ABCDE\) with \(A\) on its circumcircle.
   - We need to find the distance from \(A\) to the line \(BE\).
   - Let \(d\) be the distance from \(A\) to the line \(BE\).

5. **Use the distances \(a\), \(b\), and \(c\) to find \(d\):**
   - By the lemma, the product of the distances from \(A\) to opposite sides of the cyclic quadrilateral is equal.
   - Therefore, \(a \cdot c = b \cdot d\).

6. **Solve for \(d\):**
   \[
   d = \frac{a \cdot c}{b}
   \]

The final answer is \(\boxed{\frac{a \cdot c}{b}}\).