<s>[INST] Find the length of chord $BC$. 
On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.

$a.)7.5\quad cm\\
\color{green}{b.)7\quad cm}\\
c.)7.75\quad cm\\
d.)\text{cannot be determined}$

I constructed $BO$ and with the help of cosine rule i found $\angle AOB$.then i found $\angle BOC$
and then again applying cosine rule in $\triangle BOC$ i found $BC$ ,
but i m looking for a more simple short way.
I have studied maths up to $12th$ grade.
 [/INST] Drop altitudes from $B$ and $C$ to $AD$, and call them $X, Y$ respectively.
Then, by symmetry, $AX = DY = 4 - \frac{BC}{2}$.
Furthermore $OX = \frac{BC}{2}$.
Also, $OB = 4$ since $AD = 8$.
Now use pythagorean theorem twice:
$AX^2 + BX^2 = AB^2$
$OX^2 + BX^2 = OB^2$
$OX^2 - AX^2 = OB^2 - AB^2$
$\frac{1}{4}BC^2 - (4 - \frac{BC}{2})^2 = 12$
Solving gives $BC = 7$.
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