Law of Sine proof Below is the Law of Sines formula:
$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

And the proof is derived along the lines of:
$\sin B = \frac{h}{c}$ & $\sin C = \frac{h}{b}$ 
$c \sin B = h$ & $b \sin C = h$
$c \sin B = b \sin C$
and so on ..
Mathematically, I can understand this formula, but visually, it makes no intuitive sense to me.
How can "$c \sin B$" be equal to "$b \sin C$" when visually they have completely different proportions ?
 A: Maybe it's more intuitive this way: let $D$ be the point where the altitude meets $BC$.  Then $\triangle ABD$ and $\triangle CDA$ are both similar to $\triangle CBA$.  In particular, $\angle B = \angle DAC$.
Now $\sin\angle B = \frac{h}{c}$ and $\cos\angle DAC = \frac{h}{b}$  So
$$
     c \sin\angle B = b \cos\angle DAC
$$
But the co-sine of an angle is the sine of its co-mplement.  So $\cos\angle DAC = \sin\angle C$, which means that
$$
    c \sin\angle B = b \cos\sin\angle C
$$
whence
$$
    \frac{c}{\sin\angle C} = \frac{b}{\sin\angle B}
$$
A: I'm sorry to tell, that there's nothing I can do in order to make that proof more intuitive. It sometimes just happens that equations might be mathematically correct, but might not be intuitive (like it usually happens, for example, in probability).
The only thing I can do to help you, is to provide another proof (which - in the best case - might help you visualize the Law of Sines).

How does this picture help? 
We are given a triangle $\triangle ABC$ inscribed in a circle. We want to find out $\sin(\angle BAC)$. Thus, we take $A$ to $A'$, such that $B, O, A'$ are colinear. Observe now that $$\angle BAC=\angle BA'C\implies \sin(\angle BAC)=\sin(\angle BA'C)=\frac a{2R}$$ Through permutation, we find out that 
$$\sin\alpha=\frac a{2R}\qquad\sin\beta=\frac b{2R}\qquad\sin\gamma=\frac c{2R}$$ Or 

$$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac c{\sin\gamma}=2R$$

