Recently I have interested in the law of sines in higher dimension, so I found the result from Wikipedia with the link below:
https://en.wikipedia.org/wiki/Law_of_sines#Higher_dimensions
The statement is as follows:
Let $[v_{0},\cdots,v_{n}]$ be an $n-$simplex with vertices $v_{i}\in\mathbb{R}^n$, and we denote $|[v_{0},\cdots,v_{n}]|$ as the volume of $n-$simplex, then $$\dfrac{|[v_{1},\cdots,v_{n}]|}{psin(v_{0},v_{1}\cdots v_{n})}=\dfrac{|[v_{0},v_{2},\cdots,v_{n}]|}{psin(v_{1},v_{0}v_{2}\cdots v_{n})}=\cdots=\dfrac{|[v_{0},\cdots,v_{n-1}]|}{psin(v_{n},v_{0}\cdots v_{n-1})}$$ where $psin(v_{0},v_{1}\cdots v_{n})$ is called n-dimensional polar sine which is defined by $$psin(v_{0},v_{1}\cdots v_{n})=\dfrac{\det(v_{1}-v_{0},\cdots,v_{n}-v_{0})}{\|v_{1}-v_{0}\|\cdots\|v_{n}-v_{0}\|}$$
My definition of polar sine would be a little bit different with the Wiki's, because $v_{0},\cdots,v_{n}$ are points but not vectors here.
Meanwhile, I found a article about this theorem including the proof:
https://link.springer.com/article/10.1007/BF00181352
However, the statement is totally different:
Let $[v_{0},\cdots,v_{n}]$ be an $n-$simplex with vertices $v_{i}\in\mathbb{R}^n$, and we denote $|[v_{0},\cdots,v_{n}]|$ as the volume of $n-$simplex, then $$\dfrac{|[v_{1},\cdots,v_{n}]|}{^nsin(v_{0},v_{1}\cdots v_{n})}=\dfrac{|[v_{0},v_{2},\cdots,v_{n}]|}{^nsin(v_{1},v_{0}v_{2}\cdots v_{n})}=\cdots=\dfrac{|[v_{0},\cdots,v_{n-1}]|}{^nsin(v_{n},v_{0}\cdots v_{n-1})}$$ where $^nsin(v_{0},v_{1}\cdots v_{n})$ is called n-dimensional sine which defined by (For convenience, denote $u_{i}=v_{i}-v_{0}$) $$^nsin(v_{0},v_{1}\cdots v_{n})=\dfrac{|\det(u_{1},\cdots,u_{n})|^{n-1}}{|\det(u_{2},\cdots,u_{n})||\det(u_{1},u_{3},\cdots,u_{n})|\cdots|\det(u_{1},\cdots,u_{n-1})|}$$
Note that my definition of n-dimensional sine would be little bit different with the article. The notation $|[v_{1},\cdots,v_{n}]|$ in the article is actually the volume of parallelotope with sides (vector) $v_{1},\cdots,v_{n}$, hence it's same as $\det(v_{1},\cdots,v_{n})$.
I didn't find any mistake in this proof, so I think it's the right argument. Nevertheless these two arguments are different since the difference between n-dimensional sine and n-dimensional polar sine. It's easily to see that they're same when $n=2$. However, if $n=3$, i.e., let $ABCD$ be a tetrahedron. Then
$$psin(A,BCD)=\dfrac{6V}{|\overrightarrow{AB}||\overrightarrow{AC}||\overrightarrow{AD}|}$$
$$^nsin(A,BCD)=\dfrac{36V^2}{2A(\Delta ABC)A(\Delta ABD)A(\Delta ACD)}$$
obviously they are different. Moreover, if $\dfrac{psin(A,BCD)}{psin(B,ACD)}=\dfrac{^nsin(A,BCD)}{^nsin(B,ACD)}$, then it would be ok. However,
$$\dfrac{psin(A,BCD)}{psin(B,ACD)}=\dfrac{\overline{BC}\cdot \overline{BD}\cdot \overline{BA}}{\overline{AB}\cdot\overline{AC}\cdot\overline{AD}}=\dfrac{\overline{BC}\cdot\overline{BD}}{\overline{AC}\cdot\overline{AD}}$$
$$\dfrac{^nsin(A,BCD)}{^nsin(B,ACD)}=\dfrac{A(BCD)\cdot A(BCA)\cdot A(BDA)}{A(ABC)\cdot A(ABD)\cdot A(ACD)}=\dfrac{A(BCD)}{A(ACD)}$$
which may be different. Therefore I was confused about which version is correct, or maybe I just misunderstanding something. I need some advice, thanks!