4. On the base $AB$ of the isosceles trapezoid $ABCD$, a point $P$ is chosen, dividing the base in the ratio $AP: BP = 4: 1$. It turns out that $\angle CPD = \angle PAD$. Find the ratio $PD / PC$.
4. Answer. 2:1. Solution. Draw segments $P C$ and $P D$. By the property of parallel lines, $\angle P C D = \angle C P B$ and $\angle C D P = \angle A P D$. Then, taking into account the condition, we obtain the similarity of triangles $C P D$ and $A P D$ and the proportion $P D / P C = A P / A D$. Similarly, due to the isosceles property of the trapezoid, we obtain the similarity of triangles $C P D$ and $B P C$ and the proportion $P D / P C = B C / P B$. Multiplying these proportions and using the equality $A D = B C$, we get $(P D / P C)^{2} = A P / B P \Rightarrow P D / P C = 2$. Evaluation criteria. Correct solution - 7 points. The similarity of two pairs of triangles is proven, but there is no further progress - 2 points. The same is done only for one pair of triangles - 1 point. In all other cases - 0 points. Note. The proof does not depend on which of the bases point P is on. Therefore, it is not necessary to consider two cases.