<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] 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 $BC=7$
</s>