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.
 A: 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 $BC = 7$.
A: Drop the perpendicular from $O$ to $BC$ at $R$ call the length $b$. Drop a perpendicular on $AD$ from $B$ at the point $P$, it has length $b$ since $BC$ is parallel to $AD$. Let the length of $BC$ be $2a$ so the length RB is $a$. Look at the triangle $ \triangle ABR$, by Pythagoras we get $AB^2= b^2 + (AO-a)^2$. Since $AO$ has length 4, we have $2^2 = b^2 + (4-a)^2$. Simplifying we get $a^2+b^2 -8a= - 12$. Now look at the triangle $OPB$, by Pythagoras we get $a^2+b^2=4^2$, we substitute this value in the previous equation we get $8a=28$ or $a=7/2$, so the length of $BC$ is $2a=7$ centimeters.
A: Calculate $BD=2\sqrt{15}$ via Pythagoras.  From here you know the distance of the parallel lines $2BD/8=\sqrt{15}/2$.  Apply Pythagoras again: $4=15/4+(8-BC/2)^2$.
A: Join the points B & D to obtain right $\Delta ABD$. Thus, we get $$BD=\sqrt{(AD)^2-(AB)^2}=\sqrt{8^2-2^2}=2\sqrt{15}$$ Now, draw a perpendicular say BM from the point B to the hypotenuse AD in right $\Delta ABD$.  Then the normal distance between the parallel chords BC & AD is equal to BM i.e. length of perpendicular drawn from the vertex B to the hypotenuse AD of right $\Delta ABD$
$$BM=\frac{(AB)\times(BD)}{\sqrt{(AB)^2+(BD)^2}}=\frac{(2)\times(2\sqrt{15})}{\sqrt{(2)^2+(2\sqrt{15})^2}}=\frac{4\sqrt{15}}{8}=\frac{\sqrt{15}}{2}$$  
Let the length of BC be $x$. Now, consider a right $\Delta AMB$ with hypotenuse $AB=2 cm$ & legs $BM=\frac{\sqrt{15}}{2}$ & $$AM=\frac{AD-BC}{2}=\frac{8-x}{2}$$ Applying pythagorus theorem in right $\Delta AMB$ as follows $$(AB)^2=(BM)^2+(AM)^2 \implies (2)^2=\left(\frac{\sqrt{15}}{2}\right)^2+\left(\frac{8-x}{2}\right)^2 \implies 16=15+(8-x)^2$$ $$\implies 8-x=\pm 1 \implies x=7 \space \text{or} \space x=9$$ But $BC<AD$ hence, we have $BC=7$
