Wikipedia:Reference desk/Archives/Mathematics/2017 December 9

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December 9

Sines in a triangle

Let ABC be a triangle and O a point in the interior, then sin ∠BAO sin ∠CBO sin ∠ACO = sin ∠OAC sin ∠OBA sin ∠OCB. This isn't hard to prove being an application of the Law of Sines, and I assume it's well known, so does anyone know a reference for it? --RDBury (talk) 17:33, 9 December 2017 (UTC)

Have you looked at Law of sines? Dolphin (t) 20:46, 9 December 2017 (UTC)

That article seems to only cover a simple triangle, but what I'm looking for is more a relation between the angles in a complete quadrangle. Besides, the Law of Sines is certainly not the only way to prove the relation. It seems to be similar to Ceva's theorem but using sines, I don't see the exact connection though. --RDBury (talk) 23:59, 9 December 2017 (UTC)

Ok, now I see the connection to Ceva's theorem. In fact, the relations seems to be known as the 'trigonometric form of Ceva's theorem'. I found a reference on p66 (equation 10) of Proceedings of the Edinburgh Mathematical Society vol XX (1901-1902) but it's not given a name there. I'd still be interested to know if there is an earlier or more definitive reference though. --RDBury (talk) 00:56, 10 December 2017 (UTC)

@RDBury: The article indeed covers a single triangle, but we have just three single triangles here. Consider the ABO triangle: ${\tfrac {|BO|}{\sin \angle BAO}}={\tfrac {|AO|}{\sin \angle OBA}}$, which implies $\sin \angle BAO=\sin \angle OBA{\tfrac {|BO|}{|AO|}}$. Similary from triangles BCO and CAO we have $\sin \angle CBO=\sin \angle OCB{\tfrac {|CO|}{|BO|}}$ and $\sin \angle ACO=\sin \angle OAC{\tfrac {|AO|}{|CO|}}$. Multiply all three equalities side-wise ad we get:
$\sin \angle BAO\cdot \sin \angle CBO\cdot \sin \angle ACO=\sin \angle OBA{\tfrac {|BO|}{|AO|}}\cdot \sin \angle OCB{\tfrac {|CO|}{|BO|}}\cdot \sin \angle OAC{\tfrac {|AO|}{|CO|}}$
All line segments' lengths will get reduced and the claimed equality gets proven. --CiaPan (talk) 21:19, 10 December 2017 (UTC)

That's pretty much the proof I had in mind, but it's not covered in the article other than you're using the theorem that the article is about. Like I said, it's not hard to prove but you need to say more than just "By the law of sines." The problem I was having was finding a reference since it was too easy not to be already known. --RDBury (talk) 23:42, 10 December 2017 (UTC)

@RDBury: Oh, I see. I apologise, I must have misunderstood you. It was obvious to me that the Dolphin's comment was just a hint, and I was surprised you seemingly didn't get the hint and expect the page contains an exhaustive answer to the problem. :) --CiaPan (talk) 06:41, 11 December 2017 (UTC)