Problem - 271
In the diagram, $AB$ is the diameter of the semicircle, $\angle{CAB} = 45^\circ$, $E$ is the midpoint of $AC$, and $DE \parallel AB$. Find $\angle{ACD}$ in degrees.
Answer
15
Connect $CO$ and $DO$. Because $E$ is the middle point of $AC$ and $DE\parallel AB$, we have $DE$ is perpendicular to $OC$, and also bisect it. This means $CD=OD$. Because $DO$ is a radius, therefore it must equal to $CO$. It follows that $\triangle{CDO}$ is an equilateral triangle and $\angle{DCO}=60^{\circ}$ and $x=\boxed{15^{\circ}}$.