Three points $A,B,C$ are such that $B\in AC$. On one side of $AC$, draw the three semicircles with diameters $AB,BC,CA$. The common interior tangent at $B$ to the first two semicircles meets the third circle $E$. Let $U,V$ be the points of contact of the common exterior tangent to the first two semicircles.
Evaluate the ratio $R=\frac{[EUV]}{[EAC]}$ as a function of $r_{1} = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$, where $[X]$ denotes the area of polygon $X$.
1. **Given Information and Setup:**
   - Points \( A, B, C \) are collinear with \( B \in AC \).
   - Semicircles are drawn on one side of \( AC \) with diameters \( AB, BC, \) and \( CA \).
   - The common interior tangent at \( B \) to the semicircles with diameters \( AB \) and \( BC \) meets the third semicircle at point \( E \).
   - \( U \) and \( V \) are the points of contact of the common exterior tangent to the semicircles with diameters \( AB \) and \( BC \).

2. **Using the Pythagorean Theorem:**
   - For \( \triangle AEB \):
     \[
     EB^2 + (2r_1)^2 = AE^2
     \]
   - For \( \triangle EBC \):
     \[
     EB^2 + (2r_2)^2 = EC^2
     \]

3. **Combining the Equations:**
   - Adding the two equations:
     \[
     2EB^2 + 4r_1^2 + 4r_2^2 = AE^2 + EC^2
     \]
   - Since \( AE^2 + EC^2 = AC^2 \) and \( AC = 2(r_1 + r_2) \):
     \[
     2EB^2 + 4r_1^2 + 4r_2^2 = (2r_1 + 2r_2)^2
     \]
   - Simplifying:
     \[
     2EB^2 + 4r_1^2 + 4r_2^2 = 4(r_1 + r_2)^2
     \]
     \[
     2EB^2 + 4r_1^2 + 4r_2^2 = 4r_1^2 + 8r_1r_2 + 4r_2^2
     \]
     \[
     2EB^2 = 8r_1r_2
     \]
     \[
     EB^2 = 4r_1r_2
     \]
     \[
     EB = 2\sqrt{r_1r_2}
     \]

4. **Finding \( AE \) and \( EC \):**
   - For \( AE \):
     \[
     AE^2 = EB^2 + (2r_1)^2 = 4r_1r_2 + 4r_1^2
     \]
     \[
     AE = 2\sqrt{r_1(r_1 + r_2)}
     \]
   - For \( EC \):
     \[
     EC^2 = EB^2 + (2r_2)^2 = 4r_1r_2 + 4r_2^2
     \]
     \[
     EC = 2\sqrt{r_2(r_1 + r_2)}
     \]

5. **Area of \( \triangle AEC \):**
   - Using the formula for the area of a triangle:
     \[
     [AEC] = \frac{1}{2} \times AC \times \text{height from } E
     \]
     - The height from \( E \) is \( 2\sqrt{r_1r_2} \):
     \[
     [AEC] = 2(r_1 + r_2) \times \sqrt{r_1r_2}
     \]

6. **Finding \( EV \) and \( UE \):**
   - For \( \triangle AUB \sim \triangle AEC \):
     \[
     \frac{UB}{EC} = \frac{2r_1}{2(r_1 + r_2)}
     \]
     \[
     EV = UB = 2\sqrt{r_2(r_1 + r_2)} \cdot \frac{r_1}{r_1 + r_2}
     \]
   - For \( \triangle BVC \sim \triangle AEC \):
     \[
     \frac{BV}{AE} = \frac{2r_2}{2(r_1 + r_2)}
     \]
     \[
     UE = BV = 2\sqrt{r_1(r_1 + r_2)} \cdot \frac{r_2}{r_1 + r_2}
     \]

7. **Area of \( \triangle EUV \):**
   - Using the formula for the area of a triangle:
     \[
     [EUV] = \frac{1}{2} \times EV \times UE
     \]
     \[
     [EUV] = \frac{1}{2} \times 2\sqrt{r_2(r_1 + r_2)} \cdot \frac{r_1}{r_1 + r_2} \times 2\sqrt{r_1(r_1 + r_2)} \cdot \frac{r_2}{r_1 + r_2}
     \]
     \[
     [EUV] = \frac{2r_1r_2\sqrt{r_1r_2}}{r_1 + r_2}
     \]

8. **Ratio \( R \):**
   \[
   R = \frac{[EUV]}{[EAC]} = \frac{\frac{2r_1r_2\sqrt{r_1r_2}}{r_1 + r_2}}{2(r_1 + r_2)\sqrt{r_1r_2}}
   \]
   \[
   R = \frac{r_1r_2}{(r_1 + r_2)^2}
   \]

The final answer is \(\boxed{\frac{r_1r_2}{(r_1 + r_2)^2}}\).