[u]level 1[/u]


[b]p1.[/b] Construct the given figure, where $ABCD$ is a square and $AEF$ is an equilateral triangle.
[img]https://cdn.artofproblemsolving.com/attachments/f/c/1b4043aeed5992ddb8739eec5a8e72ebf4cf91.gif[/img]


[b]p2. [/b]Let $ABC$ be an isosceles triangle ($AB = AC$). We draw the perpendicular bisector $m$ of $AC$ and the bisector $n$ of angle $\angle C$. If $m, n$ and $AB$ intersect at a single point, how much is angle $\angle A$?


[b]p3.[/b] Let $ A$, $ B$, and $C$ be points on a circle. Let us call the orthocenter of the triangle $H$. Find the locus of $H$ as $A$ moves around the circle.

 

[u]level 2[/u]


[b]p4.[/b] Given $3$ points $ A$, $ B$ and $C$, construct the isosceles trapezoid $ABCD$ where $AB = CD$ and $BC$ is parallel to $AD$ ($BC$ different from $AD$).


[b]p5.[/b] Let $ A$, $ B$ and $C$ be points on a circle. Let's call the centroid of the triangle $G$. Find the locus of $G$ as $A$ moves along the circle.


[b]p6. [/b] Given a triangle $ABC$, let $D$, $E$, and $F$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively. From $D$ the lines $M_1$ and $M_2$ are drawn, perpendicular on $AB$ and $AC$ respectively. From $E$ the lines $M_3$ and $M_4$ are drawn, perpendicular on $BC$ and $AB$ respectively. From $F$ the lines $M_5$ and $M_6$ are drawn perpendicular on $AC$ and $BC$ respectively. Let $A'$ be the intersection between $M_4$ and $M_5$. Let $B'$ be the intersection between $M_6$ and $M_1$. Let $C'$ be the intersection between $M_2$ and $M_3$. Show that the triangles $ABC$ and $A'B'C'$ are similar and find the ratio of similarity.
To show that triangles $ABC$ and $A'B'C'$ are similar and to find the ratio of similarity, we will proceed step-by-step.

1. **Identify Midpoints and Perpendiculars:**
   - Let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$ respectively.
   - From $D$, draw lines $M_1$ and $M_2$ perpendicular to $AB$ and $AC$ respectively.
   - From $E$, draw lines $M_3$ and $M_4$ perpendicular to $BC$ and $AB$ respectively.
   - From $F$, draw lines $M_5$ and $M_6$ perpendicular to $AC$ and $BC$ respectively.

2. **Intersection Points:**
   - Let $A'$ be the intersection of $M_4$ and $M_5$.
   - Let $B'$ be the intersection of $M_6$ and $M_1$.
   - Let $C' be the intersection of $M_2$ and $M_3$.

3. **Prove Similarity:**
   - We need to show that $\triangle ABC \sim \triangle A'B'C'$.
   - Since $D$, $E$, and $F$ are midpoints, $AD$, $BE$, and $CF$ are medians of $\triangle ABC$.
   - The lines $M_1$, $M_2$, $M_3$, $M_4$, $M_5$, and $M_6$ are perpendicular to the sides of $\triangle ABC$ and pass through the midpoints of the sides.
   - By construction, $A'$, $B'$, and $C'$ are formed by the intersections of these perpendiculars.

4. **Use Midline Theorem:**
   - The midline theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.
   - Therefore, $A'B' \parallel AB$, $B'C' \parallel BC$, and $C'A' \parallel CA$.
   - Also, $A'B' = \frac{1}{2} AB$, $B'C' = \frac{1}{2} BC$, and $C'A' = \frac{1}{2} CA$.

5. **Conclude Similarity:**
   - Since $A'B' \parallel AB$, $B'C' \parallel BC$, and $C'A' \parallel CA$, and the corresponding sides are in the ratio $1:2$, $\triangle ABC \sim \triangle A'B'C'$ by the AA (Angle-Angle) similarity criterion.

6. **Ratio of Similarity:**
   - The ratio of similarity is $\frac{A'B'}{AB} = \frac{1}{2}$.

The final answer is $\boxed{\frac{1}{2}}$.