[u]level 1[/u]


[b]p1.[/b] Given an isosceles triangle $ABC$ with $AB = BC$ and $\angle ABC = 60^o$. Let $D$ in $AB$ such that $\angle DCA = 30^o$. Let $E$ in $BC$ such that $\angle EAC = 40^o$. Let $P$ be the intersection between $DC$ and $AE$. Find the measure of $\angle BPE$.


[b]p2.[/b] Given a triangle $ABC$, circles $C_1$ and $C_2$ are drawn that pass through $B$ and $C$ and intersect the sides $AB$ and $AC$ at $B ’$,$C’ $ (on $C_1$) and $B''$, $C''$ (on $C_2$). Show that $B'C'$ and $B''C''$ are parallel.


[b]p3. [/b] Given a triangle $ABC$.
i) Construct a parallelogram $DEFG$ with $E$ in $AB$, $D$ in $AC$, and$ FG$ included in $BC$ so that $BC / FG = 3$.
ii) Find the locus of the centers of the parallelograms $DEFG$, from the previous item.


[b]p4. [/b] Given an acute triangle $ABC$, let $I$ be its incenter. Let $D$ be the foot of the perpendicular from $I$ to $BC$, E the foot of the perpendicular from $I$ to $AC$, and $F$ the foot of the perpendicular from $I$ to $AB$. The circles of diameters $DI$, $EI$ and $FI$ are drawn.
i) Prove that these circles intersect in pairs, at points (other than $I$) belonging to the angle bisectors of $ABC$.
ii) Prove that the triangle formed by these intersections is similar to triangle $DEF$ and find the ratio of similarity.
1. Given that $\triangle ABC$ is isosceles with $AB = BC$ and $\angle ABC = 60^\circ$, it follows that $\triangle ABC$ is equilateral. Therefore, $AB = BC = AC$ and all internal angles are $60^\circ$.
2. Let $D$ be a point on $AB$ such that $\angle DCA = 30^\circ$. Since $\angle ACB = 60^\circ$ (as $\triangle ABC$ is equilateral), we have:
   \[
   \angle DCB = \angle ACB - \angle DCA = 60^\circ - 30^\circ = 30^\circ
   \]
3. Let $E$ be a point on $BC$ such that $\angle EAC = 40^\circ$. Since $\angle BAC = 60^\circ$ (as $\triangle ABC$ is equilateral), we have:
   \[
   \angle EAB = \angle BAC - \angle EAC = 60^\circ - 40^\circ = 20^\circ
   \]
4. Let $P$ be the intersection of $DC$ and $AE$. We need to find the measure of $\angle BPE$.
5. Consider the angles around point $P$. Since $P$ lies on both $DC$ and $AE$, we can use angle chasing to find $\angle BPE$.
6. First, note that $\angle DCA = 30^\circ$ and $\angle EAC = 40^\circ$. Therefore, $\angle DCE = \angle DCA + \angle EAC = 30^\circ + 40^\circ = 70^\circ$.
7. Since $\angle DCB = 30^\circ$ and $\angle EAB = 20^\circ$, we have:
   \[
   \angle DPE = 180^\circ - \angle DCB - \angle EAB = 180^\circ - 30^\circ - 20^\circ = 130^\circ
   \]
8. Now, consider $\angle BPE$. Since $P$ is the intersection of $DC$ and $AE$, we have:
   \[
   \angle BPE = 180^\circ - \angle DPE = 180^\circ - 130^\circ = 50^\circ
   \]

The final answer is $\boxed{50^\circ}$