B4 Consider triangle $A B C$ with angles $\angle B A C=24^{\circ}$ and $\angle A C B=28^{\circ}$. Point $D$ is constructed such that $A B$ is parallel to $C D, A D=B C$, and $A D$ and $B C$ are not parallel. Similarly, point $E$ is constructed such that $A E$ is parallel to $B C, A B=C E$, and $A B$ and $C E$ are not parallel. Lines $D E$ and $A C$ intersect at point $P$. Determine angle $\angle C P E$ (in degrees).
Answer: 4
Solution 1:
By construction, the points $A, B, C, D, E$ are all concyclic because $A B C D$ and $A B C E$ are isosceles trapezoids.
From triangle $A B C, \angle A B C=180^{\circ}-24^{\circ}-28^{\circ}=128^{\circ}$;
From trapezoid $A B C E, \angle B C E=\angle A B C=128^{\circ}$;
$\angle A C E=\angle B C E-\angle B C A=128^{\circ}-28^{\circ}-100^{\circ} ;$
$\angle D E A=\angle D C A=\angle B A C=24^{\circ}$;
From trapezoid $A B C E, \angle C E A=180^{\circ}-\angle B C E=180^{\circ}-128^{\circ}=52^{\circ}$;
$\angle D E C=\angle D E A+\angle A E C=24^{\circ}+52^{\circ}=76^{\circ}$;
Thus, from triangle $C P E, \angle C P E=180^{\circ}-100^{\circ}-76^{\circ}=4^{\circ}$.
Solution 2:
By construction, $A B C D$ and $A B C E$ are isosceles similar trapezoids. Let $F$ be the point of intersection of $A E$ and $D C$. Then $A B C F$ is a parallelogram.
From triangle $A B C, \angle A B C=180^{\circ}-24^{\circ}-28^{\circ}=128^{\circ}$;
From trapezoid $A B C E, \angle B C E=\angle A B C=128^{\circ}$;
$\angle A C E=\angle B C E-\angle B C A=128^{\circ}-28^{\circ}-100^{\circ}$;
From trapezoid $A B C E, \angle C E A=180^{\circ}-\angle B C E=180^{\circ}-128^{\circ}=52^{\circ}$;
Similarly, from trapezoid $D A B C, \angle A D C=52^{\circ}$.
Since $\angle A D C=\angle C E A$ and $\angle A F D=\angle C F E$, triangles $A F D$ and $C F E$ are similar, so $\frac{D F}{E F}=\frac{A D}{C E}$. But $A D=B C$ and $C E=A B$, so $\frac{D F}{E F}=\frac{B C}{A B}$.
Since $\angle D F E=\angle A F C=\angle A B C$, triangles $A B C$ and $E F D$ are similar.
Thus, $\angle D E A=\angle B A C=24^{\circ}$;
Therefore, $\angle D E C=\angle D E A+\angle A E C=24^{\circ}+52^{\circ}=76^{\circ}$;
Thus, from triangle $C P E, \angle C P E=180^{\circ}-100^{\circ}-76^{\circ}=4^{\circ}$.