# Proving the Sine Rule with one line.

Working on a general proof of the Law of Sines for ALL Euclidean triangles. Right triangles are easy. Acute triangles are just two proofs of the right triangle. But this is not sufficient for me. I want a one time proof for all Euclidean triangles. That brings me to the obtuse case. The first step is to draw the perpendicular to the base. This gives me two right triangles so I can prove the ratios for the opposing angles and sides. So that:

$$\frac {\sin\alpha} {a}=\frac{\sin\gamma}{c}$$

(that is the easy part)

My problem is proving that the sine of the two halves of angle $\beta$ call them $\beta_1$ and $\beta_2$ over the two halves of the bisected side (call them $b_1$ and $b_2$) satisfy the proof. So that : $$\frac {\sin \alpha}{a} = \frac{\sin\gamma}{c}= \frac {\sin(\beta_1 + \beta_2)}{b_1+b_2}$$ with out resorting to drawing more lines or using the ambiguous case.

• Have you considered using the circumcircle of a triangle along with Thales Theorem? Coxeter's "Geometry Revisited" provides a great proof of the Law of Sines using this method. amazon.com/Geometry-Revisited-New-Mathematical-Library/dp/… look at the first pages here and you'll see an elegant proof. – Tim Clark Feb 26 '15 at 21:12
• what about getting from the idea that twice the area of the triangle is the product of two sides and the $\sin$ of the angle between them? – abel Feb 26 '15 at 22:11
• @Chris: Not to lessen your enthusiasm for abel's contribution, but ... the "SAS" proof is pretty-much the standard proof of the Law of Sines. (I think the one Tim mentions using Thales' Theorem is "better", in that it explains that $a/\sin A$, $b/\sin B$, $c/\sin C$ are all equal to each other because they're equal to the same other thing, namely the diameter of the triangle's circumcircle. But that's beside the point.) Now I'm curious: If you haven't seen either of these proofs, then how was the Law of Sines explained to you in the first place? – Blue Mar 3 '15 at 3:49
• @Blue Any monkey can to use a hammer. Pretty much the story here. I actually stumbled on the sine rule (ie found a hammer) and have used it "on faith" until now. Now my daughter is in precal (much farther than I got in HS) So the reasoning and proof are important to her and therefore to me. – Chris Mar 3 '15 at 15:05

Draw a Circum-circle around the triangle and a diameter that passes through intersection of $a$ and $c$. Draw a cyclic quadrilateral including the endpoint of the diameter by joining lines of the triangle opposite to the given triangle.Mark angles $\alpha$ and $\gamma$ in the opposite segment as shown.It is clear that for all positions of right side vertex $\gamma+ \alpha$ which is constant on the arc
$$\frac {\sin\alpha} {a}=\frac{\sin\gamma}{c} = \frac{\sin(\beta_1+\beta_2)}{(b_1+b_2)}= \frac{1} {CircumDiameter D}$$