Find the ratio between $|AB|$ and $|BC|$ I have this problem, which is about this right triangle below. It says that $|AB|$ and $|BD|$ (which is the diameter of the circle) are equal and that the circle is touching the side $|AC|$. Now I have to determine the fraction $\dfrac{|AB|}{|BC|}$

I have tried a couple of things and I got the result $\frac{1}{\sqrt{15}}$. Is that correct? If not I will be pleased if somebody would give me some hints. 
And an important detail: I may not use trigonometry!
 A: Let $E$ be the point where the circle touches the hypotenuse. Then we have $|AE| = |AB|$. We also have $|CE|^2 = |CD|\cdot |BC| = |BC|(|BC| - |AB|)$ (this quantity is called the power of the point $C$ with respect to that circle). With all these sizes accounted for, the Pythagorean theorem gives
$$
|AB|^2 + |BC|^2 = |AC|^2\\
|AB|^2 + |BC|^2 = \left(|AB| + |CE|\right)^2\\
|AB|^2 + |BC|^2 = |AB|^2 + 2|AB||CE| + |BC|^2 - |BC||AB|\\
|BC||AB| = 2|AB||CE|\\
|BC|= 2|CE| \\
|BC|^2 =4|CE|^2 = 4|BC|(|BC| - |AB|)\\
|BC| = 4(|BC| - |AB|)\\
\frac{|AB|}{|BC|} = \frac{3}{4}
$$
A: Here's a solution which doesn't use the Pythogorean theorem:
Let $E$ be the point, as in Arthur's solution, where the semicircle touches the hypothenuse, and $O$ the center of the semicircle. Then note that the triangles $\Delta ECO$ and $\Delta BAC$ are similar, with the rate $2$, i.e. if $|OC| = x$, then $|AC| = 2x$ (note that they have the same angles and the angle $\angle BCA$ sees an edge of length $r$ in one triangle, and an edge of length $2r$ in the second).
Similarly, if we let $|BO| = r$ (i.e. $|AB| = 2r$) then we have $|EC| = (r + x)/2$, and $|AE| = 2x - (r + x)/2 = (3/2) x - r/2$.
Using the congruence of the triangles $\Delta OBA$ and $\Delta OEA$ we obtain
$$2r = (3/2) x - r/2$$
And conclude that $x = (5/3) r$. Using $|AB| = 2r$ and $|BC| = r + x$ we obtain
$$\frac{|AB|}{|BC|} = \frac{3}{4}$$
