[u]level 1[/u]


[b]p1.[/b] Given a triangle $ABC$, construct the rhombus $BEFG$ knowing that $E$, $F$ and $G$ are points belonging to each of the sides of the triangle (one per side).


[b]p2.[/b] Given a quadrilateral $ABCD$ (in that order), find the measure of $\angle BDC$ if $\angle ABD = 45^o$, $\angle DBC = 60^o$, $\angle BAC = 30^o$, $\angle CAD = 60^o$.


[b]p3.[/b] Given an equilateral triangle $ABC$, let $A^+$ be the symmetric of $A$ with respect to $B$, let $A^-$ be the symmetric of $A$ with respect to $C$. Let $B^+$ be the symmetric of $B$ with respect to $C$, let $B^-$ be the symmetric of $B$ with respect to $A$ and let $C^+$ be the symmetric of $C$ with respect to $A$, let $C^-$ be the symmetric of $C$ with respect to $B$. The lines are drawn $A^+A^-$, $B^+B^-$ and $C^+C^-$ forming the triangle $A'B'C'$ at the intersections of these lines. Prove that these triangles are similar. Find out the similarity ratio ($AB/ A'B'$).


[b]p4.[/b] Construct an isosceles trapezoid $ABCD$ ($AB \parallel CD$) knowing that its diagonals are perpendicular and that $AB = BD$.


[b]p5.[/b] Given a triangle $ABC$, and an interior point $P$, draw $r, s$, and $t$ parallel lines through $P$ to the sides $BC$, $AB$, and $CA$ respectively. Let $F$ be the intersection between $ r$ and $AB$, let $G$ be the intersection between $ r$ and $AC$, let $D$ be the intersection between $t$ and $BC$, and let $E$ be the intersection between $ s$ and $BC$.

Find the locus of the points $P$ if the areas of the parallelograms $GPDC$ and $FPEB$ are equal.


[b]p6.[/b] In the previous problem, find the locus of the points $P$ if the perimeters of the parallelograms $GPDC$ and $FPEB$ are equal.
Given the problem, we need to find the locus of the points \( P \) such that the areas of the parallelograms \( GPDC \) and \( FPEB \) are equal.

1. **Define the coordinates of the vertices of \(\triangle ABC\):**
   \[
   A(0, a), \quad B(-b, 0), \quad C(c, 0)
   \]
   Let \( P \) be an interior point with coordinates \( P(k, l) \).

2. **Determine the coordinates of the intersection points:**
   - Point \( E \) is the intersection of line \( s \) (parallel to \( AB \)) with \( BC \):
     \[
     E\left(k - \frac{bl}{a}, 0\right)
     \]
   - Point \( D \) is the intersection of line \( t \) (parallel to \( CA \)) with \( BC \):
     \[
     D\left(k + \frac{lc}{a}, 0\right)
     \]
   - Point \( F \) is the intersection of line \( r \) (parallel to \( BC \)) with \( AB \):
     \[
     F\left(\frac{bl}{a} - b, l\right)
     \]
   - Point \( G \) is the intersection of line \( r \) (parallel to \( BC \)) with \( AC \):
     \[
     G\left(c - \frac{lc}{a}, l\right)
     \]

3. **Calculate the areas of the parallelograms \( GPDC \) and \( FPEB \):**
   - The area of parallelogram \( GPDC \):
     \[
     \text{Area}(GPDC) = \left| \frac{1}{2} \left( (c - \frac{lc}{a} - k - \frac{lc}{a})(0 - l) - (0 - l)(k + \frac{lc}{a} - k) \right) \right|
     \]
     Simplifying, we get:
     \[
     \text{Area}(GPDC) = \left| \frac{1}{2} \left( (c - \frac{2lc}{a})(-l) \right) \right| = \left| \frac{1}{2} \left( -cl + \frac{2l^2c}{a} \right) \right| = \left| \frac{l}{2} \left( -c + \frac{2lc}{a} \right) \right|
     \]
   - The area of parallelogram \( FPEB \):
     \[
     \text{Area}(FPEB) = \left| \frac{1}{2} \left( (\frac{bl}{a} - b - k + \frac{bl}{a})(0 - l) - (0 - l)(k - \frac{bl}{a} - (-b)) \right) \right|
     \]
     Simplifying, we get:
     \[
     \text{Area}(FPEB) = \left| \frac{1}{2} \left( (\frac{2bl}{a} - b - k)(-l) \right) \right| = \left| \frac{1}{2} \left( -bl + \frac{2l^2b}{a} \right) \right| = \left| \frac{l}{2} \left( -b + \frac{2lb}{a} \right) \right|
     \]

4. **Set the areas equal and solve for the locus of \( P \):**
   \[
   \left| \frac{l}{2} \left( -c + \frac{2lc}{a} \right) \right| = \left| \frac{l}{2} \left( -b + \frac{2lb}{a} \right) \right|
   \]
   Simplifying, we get:
   \[
   \left| -c + \frac{2lc}{a} \right| = \left| -b + \frac{2lb}{a} \right|
   \]
   Since \( l \neq 0 \), we can divide both sides by \( l \):
   \[
   \left| -c + \frac{2lc}{a} \right| = \left| -b + \frac{2lb}{a} \right|
   \]
   This simplifies to:
   \[
   \left| -c + \frac{2lc}{a} \right| = \left| -b + \frac{2lb}{a} \right|
   \]
   Solving for \( k \) and \( l \), we get the equation of the line:
   \[
   2ax + (c - b)y + a(b - c) = 0
   \]

The final answer is \(\boxed{2ax + (c - b)y + a(b - c) = 0}\).