Let $ ABC$ be an acute-angled triangle. Let $ L$ be any line in the plane of the triangle $ ABC$. Denote by $ u$, $ v$, $ w$ the lengths of the perpendiculars to $ L$ from $ A$, $ B$, $ C$ respectively. Prove the inequality $ u^2\cdot\tan A \plus{} v^2\cdot\tan B \plus{} w^2\cdot\tan C\geq 2\cdot S$, where $ S$ is the area of the triangle $ ABC$. Determine the lines $ L$ for which equality holds.
1. **Initial Setup and Simplification**:
   - Let $ABC$ be an acute-angled triangle.
   - Let $L$ be any line in the plane of the triangle $ABC$.
   - Denote by $u$, $v$, $w$ the lengths of the perpendiculars to $L$ from $A$, $B$, $C$ respectively.
   - We need to prove the inequality \( u^2 \cdot \tan A + v^2 \cdot \tan B + w^2 \cdot \tan C \geq 2 \cdot S \), where $S$ is the area of the triangle $ABC$.

2. **Considering the Position of Line $L$**:
   - If the line $L$ lies entirely outside the triangle, we can decrease each of $u, v, w$ by sliding $L$, keeping it parallel to its initial position, until it passes through one of the vertices.
   - Therefore, we need only consider lines that pass through the interior (including the border) of the triangle.

3. **Simplifying the Problem**:
   - We can further slide $L$ in a parallel fashion and decrease the sum of $u, v, w$ until the setup is equivalent to:
     - $A$ and $B$ lie on opposite sides of $L$, and $L$ passes through $C$.
   - This is true because $u + v$ remains constant as we slide around if and only if $B$ and $C$ are on opposite sides of $L$. Meanwhile, $w$ decreases until it is zero, the minimum, which is just when $L$ is through $A$.

4. **Rewriting the Inequality**:
   - We need to show that:
     \[
     u^2 \cdot \tan A + v^2 \cdot \tan B + w^2 \cdot \tan C \geq 2 \cdot S
     \]
   - We can show it is equivalent to:
     \[
     \frac{u^2}{b \cos A} + \frac{v^2}{a \cos B} \geq c
     \]
   - If $u = b \sin(A + X)$, then $v = a \sin(B - X)$.

5. **Verifying the Inequality**:
   - We need to show:
     \[
     b \frac{\sin^2(A + X)}{\cos A} + a \frac{\sin^2(B - X)}{\cos B} \geq c
     \]
   - We will verify by checking boundary cases as well as zeros of the derivative (as a function of $X$).

6. **Finding the Derivative**:
   - The derivative is zero when:
     \[
     b \frac{\sin 2(A + X)}{\cos A} = a \frac{\sin 2(B - X)}{\cos B}
     \]
     \[
     \iff \frac{\sin 2(A + X)}{\sin 2A} = \frac{\sin 2(B - X)}{\sin 2B}
     \]
     \[
     \iff \cos 2(A + X - B) - \cos 2(A + X + B) = \cos 2(B - X - A) - \cos 2(A + B - X)
     \]
     \[
     \iff \cos 2(A + B + X) = \cos 2(A + B - X) \iff \sin 2(A + B) \sin X = 0
     \]
   - Thus, we need only check boundaries; i.e., $X = 0, \frac{\pi}{2}$; but at both, we clearly see equality attained ($a \cos B + b \cos A = c$).

7. **Conclusion**:
   - We have proven the inequality with equality when $L$ is one of the sides of the triangle.

The final answer is \( \boxed{ u^2 \cdot \tan A + v^2 \cdot \tan B + w^2 \cdot \tan C \geq 2 \cdot S } \)