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Usually, when anyone derives the Law of Sines for any triangle, one breaks the triangle into two new right triangles by drawing altitude h to side AB. Here is an example.

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The relationships of each triangle are defined: $$ ACD: \sin A = h/b \tag{1}\label{1} $$ $$ BCD: \sin B = h/a \tag{2}\label{2} $$

But alternatively in that example altitude j was drawn on side AC using triangles BEC and AEB to derive the relationship: $$ j = a \sin C = c \sin A \tag{3}\label{3} $$

Which is valid. It follows that #3 is added with #1 and #2 when the Law of Sines equation is obtained. So what is the significance of drawing line j and why does that proof work?

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The relationships of each triangle are defined: $$ ACD: \sin A = h/b \tag{1}$$ $$ BCD: \sin B = h/a \tag{2} $$

It follows from $(1)$ and $(2)$ that $\;h = b \sin A = a \sin B\;$ where the latter equality is the same as $(3)$ but written for altitude $h$ instead of $j$. Therefore, there is no special significance to that $j$, relation $(3)$ is equivalent to $(1)+(2)$, and it is in fact not needed for the proof. From $(1)$ and $(2)$ alone it follows that $\;\cfrac{a}{\sin A}=\cfrac{b}{\sin B}\;$ then by symmetry in $\;a,b,c\;$ follows the complete law of sines:

$$\cfrac{a}{\sin A}=\cfrac{b}{\sin B}=\cfrac{c}{\sin C}$$

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    That was quick dxiv. I think what I may wanted to figure out originally is why it is possible to make altitude j (at least why the mathematicians drew it in said figure). – 2017-01-18
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    @AthanMoushos Not sure I follow the question. A triangle has $3$ altitudes, and $j$ is as legitimate an altitude as is $h$. Why *they* chose to draw it in the figure, that I can't second-guess. Maybe the context around where that figure was pulled from has more clues. – 2017-01-18