Problem 1435

1. Define the vertices and midpoints:
Let $a, b, c, d$ be the complex numbers (affixes) representing the vertices of the cyclic quadrilateral $ABCD$. The midpoints of the sides $AB, BC, CD, DA$ are given by:
$$m_1 = \frac{a+b}{2}, \quad m_2 = \frac{b+c}{2}, \quad m_3 = \frac{c+d}{2}, \quad m_4 = \frac{d+a}{2}$$

2. Assume the center of the circle:
Assume the center of the circle $O$ is at the origin, i.e., $O(0)$.

3. Define the perpendiculars:
Let $\ell_i$ be the perpendicular dropped from $m_i$ to the opposite side of the quadrilateral.

4. Form a parallelogram:
Consider the quadrilateral formed by the perpendicular bisectors of $AB$ and $CD$ and the lines $\ell_1$ and $\ell_3$. This quadrilateral is a parallelogram with $O$ as one of its vertices.

5. Intersection points:
The intersection point of $\ell_1$ and $\ell_3$ has the affix:
$$m_1 + m_3 = \frac{a+b}{2} + \frac{c+d}{2} = \frac{a+b+c+d}{2}$$
Similarly, the intersection point of $\ell_2$ and $\ell_4$ has the affix:
$$m_2 + m_4 = \frac{b+c}{2} + \frac{d+a}{2} = \frac{a+b+c+d}{2}$$

6. Conclusion:
Since $m_1 + m_3 = m_2 + m_4 = \frac{a+b+c+d}{2}$, it follows that the lines $\ell_1, \ell_2, \ell_3, \ell_4$ are concurrent at the point with affix $\frac{a+b+c+d}{2}$.

$\blacksquare$