[u]level 1[/u]


[b]p1.[/b] Given a circle $\omega$ , a point $M_b$ in its interior, construct all the triangles $ABC$ inscribed in $\omega$  such that $M_b$ is the midpoint of $AC$ and $AB = AC$.


[b]p2.[/b] Given the isosceles triangle $ABC$ with $AB = BC$, let $D$ in $AB$ and $E$ in $BC$ such that $AD = BE$. Find the locus of  the midpoint $M$ of segment $DE$.


[b]p3.[/b] Given a circle $\omega$ and a point $A$ in its interior. Construct the rectangle $ABCD$, with $BC$ in $\omega$ , of ​​maximum area.


[b]p4. [/b]Using the medians of a triangle $T$ as sides, another triangle $T'$ is constructed (prove that it is always possible to form such a triangle) and using with the medians of $T'$, another triangle  $T''$ is formed. Prove that triangle $T''$ is similar to triangle $T$.
Given the isosceles triangle $ABC$ with $AB = BC$, let $D$ be a point on $AB$ and $E$ be a point on $BC$ such that $AD = BE$. We need to find the locus of the midpoint $M$ of segment $DE$.

1. **Coordinate Setup:**
   Let's place the triangle in the coordinate plane with vertices at $A(-c, 0)$, $B(0, b)$, and $C(c, 0)$. This setup ensures that $AB = BC$.

2. **Coordinates of Points $D$ and $E$:**
   Let $D$ be a point on $AB$ with coordinates $D(\lambda, \frac{b(\lambda + c)}{c})$. This is derived from the section formula, considering $D$ divides $AB$ in the ratio $\frac{\lambda + c}{c}$.

   Similarly, let $E$ be a point on $BC$ with coordinates $E(c + \lambda, -\frac{b\lambda}{c})$. This is derived from the section formula, considering $E$ divides $BC$ in the ratio $\frac{\lambda}{c}$.

3. **Midpoint Calculation:**
   The midpoint $M$ of segment $DE$ can be found using the midpoint formula:
   \[
   M = \left( \frac{\lambda + (c + \lambda)}{2}, \frac{\frac{b(\lambda + c)}{c} + \left(-\frac{b\lambda}{c}\right)}{2} \right)
   \]
   Simplifying the coordinates of $M$:
   \[
   M = \left( \frac{2\lambda + c}{2}, \frac{b(\lambda + c - \lambda)}{2c} \right)
   \]
   \[
   M = \left( \lambda + \frac{c}{2}, \frac{b \cdot c}{2c} \right)
   \]
   \[
   M = \left( \lambda + \frac{c}{2}, \frac{b}{2} \right)
   \]

4. **Locus of Midpoint $M$:**
   From the above calculation, we see that the $y$-coordinate of $M$ is always $\frac{b}{2}$, regardless of the value of $\lambda$. Therefore, the locus of the midpoint $M$ is a horizontal line at $y = \frac{b}{2}$.

\[
\boxed{y = \frac{b}{2}}
\]