7. (3 points) Given a convex pentagon $A B C D E$. Point $P$ is the intersection of $B E$ and $A C$, point $Q$ is the intersection of $C E$ and $A D$, point $O$ is the intersection of $A D$ and $B E$. It turns out that triangles $A B P$ and $D E Q$ are isosceles triangles with the angle at the vertex (exactly at the vertex, not at the base) equal to 80 degrees. Find the value of angle $A C E$, given that triangles $A P O$ and $E Q O$ are also isosceles.
Answer: $60^{\circ}$ or $105^{\circ}$

Solution: Consider triangle $A B P$. The angle $\angle A P B$ in it can be $80^{\circ}$ or $50^{\circ}$. Then the adjacent $\angle A P O$ is $100^{\circ}$ or $130^{\circ}$. Therefore, in triangle $A P O$, this is the angle at the vertex and $\angle A O P$ is $40^{\circ}$ or $25^{\circ}$.

We can make similar considerations for triangles $B Q E$ and $E Q O$, obtaining that $\angle Q O E$ is also $40^{\circ}$ or $25^{\circ}$ depending on $\angle D Q E$. But $\angle Q O E=\angle A O P$ as vertical angles, so $\angle A P B=\angle D Q E$.

$\angle A C E=360^{\circ}-\angle C P O-\angle C Q O-\angle P O Q=360^{\circ}-540^{\circ}+\angle A P O+\angle E Q O+\angle A O P$, which is $60^{\circ}$ or $105^{\circ}$.