Let $A B C$ be an isosceles triangle with $A C=B C$. The point $D$ lies on the side $A B$ such that the semicircle with diameter $[B D]$ and center $O$ is tangent to the side $A C$ in the point $P$ and intersects the side $B C$ at the point $Q$. The radius $O P$ intersects the chord $D Q$ at the point $E$ such that $5 \cdot P E=3 \cdot D E$. Find the ratio $\frac{A B}{B C}$.
We denote $O P=O D=O B=R, A C=B C=b$ and $A B=2 a$. Because $O P \perp A C$ and $D Q \perp B C$, then the right triangles $A P O$ and $B Q D$ are similar and $\varangle B D Q=\varangle A O P$. So, the triangle $D E O$ is isosceles with $D E=O E$. It follows that

$$
\frac{P E}{D E}=\frac{P E}{O E}=\frac{3}{5}
$$

Let $F$ and $G$ be the orthogonal projections of the points $E$ and $P$ respectively on the side $A B$ and $M$ is the midpoint of the side $[A B]$. The triangles $O F E, O G P, O P A$ and $C M A$ are similar. We obtain the following relations

$$
\frac{O F}{O E}=\frac{O G}{O P}=\frac{C M}{A C}=\frac{O P}{O A}
$$

But $C M=\sqrt{b^{2}-a^{2}}$ and we have $O G=\frac{R}{b} \cdot \sqrt{b^{2}-a^{2}}$. In isosceles triangle $D E O$ the point $F$ is the midpoint of the radius $D O$. So, $O F=R / 2$. By using Thales' theorem we obtain

$$
\frac{3}{5}=\frac{P E}{O E}=\frac{G F}{O F}=\frac{O G-O F}{O F}=\frac{O G}{O F}-1=2 \cdot \sqrt{1-\left(\frac{a}{b}\right)^{2}}-1
$$

From the last relations it is easy to obtain that $\frac{a}{b}=\frac{3}{5}$ and $\frac{A B}{B C}=\frac{6}{5}$.

The problem is solved.