6. In the triangle $A B C, A C=2 B C, \angle C=90^{\circ}$ and $D$ is the foot of the altitude from $C$ onto $A B$. A circle with diameter $A D$ intersects the segment $A C$ at $E$. Find $A E: E C$.
6. Answer: 4 .

Since $A D$ is a diameter, we have $\angle A E D=90^{\circ}$ so that $D E$ is parallel to $B C$. As $\triangle A D C$ is similar to $\triangle A C B$, we have $A C / A D=A B / A C$ so that $A C^{2}=A D \times A B$. Similarly, $B C^{2}=A B \times B D$. Thus $A E / E C=A D / B D=(A C / B C)^{2}=4$.