Three points $A,B,C$ are such that $B \in ]AC[$. On the side of $AC$ we draw the three semicircles with diameters $[AB], [BC]$ and $[AC]$. The common interior tangent at $B$ to the first two semi-circles meets the third circle in $E$. Let $U$ and $V$ be the points of contact of the common exterior tangent to the first two semi-circles. Denote the area of the triangle $ABC$ as $S(ABC)$. Evaluate the ratio $R=\frac{S(EUV)}{S(EAC)}$ as a function of $r_1 = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$.
1. **Identify the points and semicircles:**
   - Let $A$, $B$, and $C$ be points on a line such that $B$ is between $A$ and $C$.
   - Draw semicircles with diameters $[AB]$, $[BC]$, and $[AC]$.

2. **Common interior tangent at $B$:**
   - The common interior tangent at $B$ to the semicircles with diameters $[AB]$ and $[BC]$ meets the semicircle with diameter $[AC]$ at point $E$.

3. **Points of contact of the common exterior tangent:**
   - Let $U$ and $V$ be the points of contact of the common exterior tangent to the semicircles with diameters $[AB]$ and $[BC]$.

4. **Projections and similarity:**
   - Let $X = EA \cap (AB)$ and $Y = EC \cap (BC)$, where $(PQ)$ denotes the semicircle with diameter $PQ$.
   - $X$ and $Y$ are the projections of $B$ on $EA$ and $EB$, respectively.
   - Since $E$ is on the radical axis of $(AB)$ and $(BC)$, we have $EX \cdot EA = EY \cdot EC$.
   - This implies $\triangle EYX \sim \triangle EAC$ by the Angle-Angle (AA) similarity criterion.

5. **Tangency and similarity:**
   - $\angle BYX = \angle EXY = \angle ECA = \angle BCY$, so $XY$ is tangent to $(BC)$.
   - Similarly, $XY$ is tangent to $(AB)$.
   - Therefore, $X = U$ and $Y = V$.

6. **Area ratio calculation:**
   - The ratio of the areas of $\triangle EUV$ and $\triangle EAC$ is given by the square of the ratio of corresponding sides:
     \[
     \frac{S(EUV)}{S(EAC)} = \left(\frac{UV}{AC}\right)^2 = \left(\frac{EB}{AC}\right)^2
     \]
   - Since $E$ is the point where the common interior tangent meets the semicircle with diameter $[AC]$, $EB$ is the geometric mean of $AB$ and $BC$:
     \[
     EB = \sqrt{AB \cdot BC}
     \]
   - Therefore,
     \[
     \frac{S(EUV)}{S(EAC)} = \left(\frac{\sqrt{AB \cdot BC}}{AB + BC}\right)^2 = \frac{AB \cdot BC}{(AB + BC)^2}
     \]
   - Using $r_1 = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$, we get:
     \[
     AB = 2r_1 \quad \text{and} \quad BC = 2r_2
     \]
   - Substituting these into the ratio:
     \[
     \frac{S(EUV)}{S(EAC)} = \frac{(2r_1) \cdot (2r_2)}{(2r_1 + 2r_2)^2} = \frac{4r_1 r_2}{4(r_1 + r_2)^2} = \frac{r_1 r_2}{(r_1 + r_2)^2}
     \]

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