In the $xOy$ system consider the lines $d_1\ :\ 2x-y-2=0,\ d_2\ :\ x+y-4=0,\ d_3\ :\ y=2$ and $d_4\ :\ x-4y+3=0$. Find the vertices of the triangles whom medians are $d_1,d_2,d_3$ and $d_4$ is one of their altitudes.

[i]Lucian Dragomir[/i]
1. **Identify the vertices on the given lines:**
   - For line \(d_1: 2x - y - 2 = 0\), let \(A(p, 2p - 2)\).
   - For line \(d_2: x + y - 4 = 0\), let \(B(4 - q, q)\).
   - For line \(d_3: y = 2\), let \(C(r, 2)\).

2. **Find the centroid of \(\triangle ABC\):**
   - The centroid \(G\) of \(\triangle ABC\) is given by:
     \[
     G = \left(\frac{p + (4 - q) + r}{3}, \frac{(2p - 2) + q + 2}{3}\right)
     \]
   - Given that \(d_1, d_2, d_3\) are medians, they must intersect at the centroid. Since \(d_1, d_2, d_3\) are concurrent at \((2, 2)\), we have:
     \[
     \left(\frac{p + (4 - q) + r}{3}, \frac{(2p - 2) + q + 2}{3}\right) = (2, 2)
     \]

3. **Solve for \(p\), \(q\), and \(r\):**
   - From the \(x\)-coordinate:
     \[
     \frac{p + (4 - q) + r}{3} = 2 \implies p + 4 - q + r = 6 \implies p - q + r = 2
     \]
   - From the \(y\)-coordinate:
     \[
     \frac{(2p - 2) + q + 2}{3} = 2 \implies 2p - 2 + q + 2 = 6 \implies 2p + q = 6
     \]

4. **Express \(q\) in terms of \(p\):**
   - From \(2p + q = 6\):
     \[
     q = 6 - 2p
     \]

5. **Substitute \(q\) into the first equation:**
   - Substitute \(q = 6 - 2p\) into \(p - q + r = 2\):
     \[
     p - (6 - 2p) + r = 2 \implies p - 6 + 2p + r = 2 \implies 3p + r = 8 \implies r = 8 - 3p
     \]

6. **Verify the vertices:**
   - \(A(p, 2p - 2)\)
   - \(B(4 - (6 - 2p), 6 - 2p) = (2p - 2, 6 - 2p)\)
   - \(C(8 - 3p, 2)\)

7. **Check if \(d_4\) is an altitude:**
   - Suppose \(d_4: x - 4y + 3 = 0\) is an altitude from \(C\). Then \(C\) must lie on \(d_4\):
     \[
     8 - 3p - 4 \cdot 2 + 3 = 0 \implies 8 - 3p - 8 + 3 = 0 \implies -3p + 3 = 0 \implies p = 1
     \]
   - Substitute \(p = 1\) into the coordinates of \(A\), \(B\), and \(C\):
     \[
     A(1, 0), \quad B(0, 4), \quad C(5, 2)
     \]

8. **Verify the slopes:**
   - Slope of \(AB\):
     \[
     \text{slope of } AB = \frac{4 - 0}{0 - 1} = -4
     \]
   - Slope of \(d_4\):
     \[
     \text{slope of } d_4 = 4
     \]
   - Since the product of the slopes of \(AB\) and \(d_4\) is \(-4 \cdot 4 = -16\), they are perpendicular, confirming \(d_4\) is an altitude.

The final answer is \(\boxed{A(1, 0), B(0, 4), C(5, 2)}\)