Length of chord DE 
In the given circle |AB| = 10 Units and AB || CD
AB is the diameter of the circle.
What is the length of the chord ED?
 A: Common case $\angle ABC = \alpha, \angle CDE = \beta$:

Let $DF || BE \Rightarrow |ED|=|BF|$. 
$\angle CBE = \angle EDC = \beta$ - same chord.
$\angle EBF = \angle BFD= \angle DCB= \angle ABC =\alpha$. 
As $\angle AFB=\frac{\pi}{2} \Rightarrow |FB|=|AB|\cdot \textbf{cos} \angle ABF$.
So $|ED|=|AB|\cdot \textbf{cos} (2\alpha+\beta)$.
A: By inscribed angle of a cirle, $\angle CBE = \angle CDE = 15^\circ$. That means $\angle ABE = 25^\circ$. Since $AB$ is a diameter, $\angle AEB$ is right, and thus we conclude that $\angle BAE = 90^\circ - 25^\circ = 65^\circ$.
Since $\angle BAD = 10^\circ$ by symmetry, we have that $\angle DAE = 65^\circ - 10^\circ = 55^\circ$. If we let $O$ be the center of the circle, we know then that $\angle DOE = 2\cdot\angle DAE = 110^\circ$. Your unknown side is now the last side of an isosceles triangle with two sides $5$ long, since they are radii in the circle, and with angles $110^\circ$, $35^\circ$ and $35^\circ$. That should be easer to calculate.
A: 
O is center of circle. and You must know that total angle of circle is 360 degree.
