Find chord length given diameter and two other chords Problem:
I'm asked to find the length of $m$, given the following diagram.

Note that $\overline{AC}$ = 1 and $\overline{CD} = 1$ and that $\overline{AB}$ is a diameter whose length is 4.
Attempt:
I know that $\angle ADB$ is a right angle and that if I can find the length $\overline{AD}$ then this will be solved. I also know that $\triangle AGE \sim \triangle ADB$. And I now that the measure of $\angle ABD$ is half the measure of angle $\angle AED$ (one segment of which is not shown).
Question:
How can I go about finding the lengths of $\overline{AG}$ or $\overline{AD}$?
 A: 
We know the following: 


*

*$\triangle ADB$ is a right triangle (by the Inscribed Right Angle Theorem)

*$t \parallel k$ (by the Corresponding Angle Theorem)

*$\triangle AFE$ is a right triangle (by the Side Angle Side Similarity Theorem)

*$\triangle CFA$ is a right triangle (by the Vertical Angle Theorem)

*$\triangle CFD$ is a right triangle (by the Vertical Angle Theorem)

*$\triangle ADB \sim \triangle AFE$ (by the Side Angle Side Similarity Theorem)


We will use the Pythagorean Theorem and the following relationships to find $k$:


*

*$s + t = 2$ (since $s + t$ equals the radius of the semicircle)

*$s^2 + q^2 = 1^2$ (since $\triangle CFA$ is a right triangle)

*$q^2 + t^2 = 2^2$ (since $\triangle AFE$ is a right triangle)

*$2t = k$ (since $\triangle ADB \sim \triangle AFE$ by a factor of 2)


Based on the relationships above, we can form a system of equations to find the value of $k$:
\begin{align*}
    (2-t)^2 + q ^2 &= 1 && \text{substitution}\\
    q^2 &= 1- (2-t)^2 \\
    q &= \sqrt{1-(2-t)^2} \\
    (\sqrt{1-(2-t)})^2 + t^2 &= 2^2 && \text{substitution}\\
    -t^2 +4t -3 + t^2 = 4 \\
    4t-3 &= 4 \\
    t &= \frac{7}{4} \\
    2(\tfrac{7}{4}) &= k && \text{substitution}\\
    k &= \frac{14}{4} \\
    k &= 3.5
\end{align*}
Thus, the measure of $k$ is $3.5$ units.
A:  To find AG or AD draw the hight of triangle ACD to cross the side AD  of triangle ABD at E.Also connect B to C. Triangle ACE  is similar to triangle ABC for :
$\angle ABC=\angle CDA=arc AC/2$
So we have:
$\frac{CE}{AC}=\frac{AC}{AB}=\frac{1}{4}$ ⇒ $CE=\frac{1}{4}$
In right triangle ACE we have:
$AE^2=AC^2-CE^2=1^2-(\frac{1}{4})^2$ ⇒ $AE=\frac{\sqrt {15}}{4}$
⇒ $AD=2\times AE=\frac{\sqrt {15}}{2}$
If G is the projection of C on AB, then two triangles ACE and ACG are equal so we have :
$AG=CE=\frac{1}{4}$
