# Sine rule - strictly geometric proof [closed]

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.

## closed as unclear what you're asking by Harish Chandra Rajpoot, egreg, Leucippus, Pragabhava, colormegoneJan 27 '16 at 0:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Yes, I mean that equation. – sZpak Jan 26 '16 at 21:49
• I think you'll have to allow proportion, since the law of sines is given as a series of proportions... – Brian Tung Jan 26 '16 at 22:06

$\triangle ABC$ 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 $BC$.
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.