Prove that the area of a triangle is $\frac12 |a \times b| = \frac12 |b \times c| = \frac12 |c \times a|$ I have a triangle $ABC$ where $\overrightarrow{AB} = \vec c, \overrightarrow{BC} = \vec a, \overrightarrow{CA} = \vec b$ and $\angle (\vec b, \vec c) = \alpha, \angle (\vec c, \vec a) = \beta, \angle (\vec a, \vec b) = \gamma$
I have to prove that the area of this triangle is given by $ \frac 12 |a \times b| = \frac 12 |b \times c| = \frac 12 |c \times a|$
I'm not sure where to begin but I don't think the following can be considered as a proof.  $\frac 12 |a \times b| = |a||b| \sin \gamma = |b||c| \sin \alpha = \frac 12 |b \times c|$
 A: Since we are talking about triangles, I believe that the sine definition of a cross product is valid in this situation.
Suppose a triangle $\triangle ABC$ has side lengths $a,b$, and $c$ with angles $A,B,$ and $C$ opposite of each respective side $a, b, $ and $c$.
By the Law of Sines: $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} $$
The cross product of two vectors $\vec{x}, \vec{y}$ in $\mathbb{R}^2$ is $$\vec{x}\times \vec{y}=||\vec{x}|| \cdot ||\vec{y}||\cdot \sin\theta $$ where $\theta$ is the angle between the two vectors.
Area of $\triangle ABC$ is $$\frac{1}{2}ab\sin C$$
I suggest drawing a picture and manipulating the Law of Sines to see why they are equal and thinking of the sides of the triangle as two dimensional vectors.
Edit- Consider the following diagram:

We know that $\frac{1}{2}ab\sin C$ is the area of the triangle. By the Law of Sines, we can obtain $$b\sin A=a\sin B$$ $$c\sin B=b\sin C$$ $$c\sin A=a\sin C $$
We also know, if we let the sides of our triangle be vectors, that $$|a\times b| = ab\sin C$$ $$|a\times c|=ac\sin B $$ $$|b\times c|=bc\sin A$$
All that is left for you to do now is to just manipulate some of the above expressions and substitute some things using Area of $\triangle ABC =\frac{1}{2}ab\sin C$.
