Sine rule - strictly geometric proof I'm looking for a proof of a sine rule without using algebra at all.
There are many proofs which uses geometric approach on the beginning but at the end there is always something like:
"now multiply both sides by x" or 
"solve for h and that gives you something".
I'm looking for something which is based only on length, area and maybe proportion...
Thanks.
------ edit ------
Ok. I've got it. Answers given by Brian Tung and Mark Bennet are fine, I can agree that they are kind of "purely geometric", but what I meant is something like that: 

It's still difficult to explain what I'm really asking for, but my intuition is that it's easier to understand geometric proof when I deal with length and area using algebra only if necessary.
 A: Take $O$ to be the circumcentre of the triangle, and join $OA. OB, OC$. Put down also the perpendiculars from the circumcentre to the three sides of the triangle. Use that the angle at the centre of the circle is twice the angle at the circumference and that the perpendiculars bisect isosceles triangles e.g. in $\triangle OAB$, $OA=OB=R$. This gives $R\sin A=\frac a2$ or $2R=\frac a{\sin A}$ simply by the definition of sine in a right-angled triangle. The ratios are all equal to the same thing.
The angle at centre = twice angle at circumference is easily proved by "pure geometry" again involving isosceles triangles.
A: 
$\triangle ABC$ is inscribed in unit circle $O$.  Bisect $\overline{AC}$ at $D$, and bisect $\overline{BC}$ at $E$.  Then $m\angle AOD$ is half of $m\angle AOC$, and also $m\angle ABC$ is half of $m\angle AOC$, so angles $\angle AOD$ and $\angle ABC$ are congruent, so their sines are equal, and their value is half the length of $\overline{AC}$.
Similarly, $m\angle BOE$ is half of $m\angle BOC$, and also $m\angle BAC$ is half of $m\angle BOC$, so angles $\angle BOE$ and $\angle BAC$ are congruent, and their sines are both equal to half the length of $\overline{BC}$.
Hence, $\sin\angle ABC$ is to $AC$ as $\sin\angle BAC$ is to $BC$.
