How is the area of an arbitrary triangle and area of a parallelogram related?

Question

In am reading a vector algebra book, it says

Theorem x.x says magnitude of the
cross product P × Q is equal to the area of the parallelogram whose sides are
formed by the vectors P and Q. As a consequence, the area A of an arbitrary tri-
angle whose vertices are given by the points V1, V2, and V3 can be calculated us-
ing formula 
-- Snippet from the book

$$A = \frac{1}{2}||(v_{2} - v_{1})X(v_{3}-v_{1})||$$

this piece of information is not even related to the vector product topic I am studying. I cannot understand how the area of an arbitrary triangle is dependent on the $PXQ$ value we've just derived?

Can some one explain to me what the above block quote line means?

Answer 1

Given a triangle $\triangle ABC$, draw $CD$ and $BD$ parallel to $AB$ and $AC$ respectively, giving a parallelogram $ABDC$. The area of $\triangle ABC$ is clearly one half the area of the parallelogram, so it is $$\frac{1}{2}\lVert AB\times AC\rVert.$$

Answer 2

Let us make sense out of $$A = \frac{1}{2}||(v_{2} - v_{1})(v_{3}-v_{1})||$$First, we notice that $ v_{2} - v_{1}$ and $v_{3}-v_{1}$ are vectors.

The norm of the cross product of these vectors gives us the area of the parallelogram whose sides are formed by these vectors. The area of the triangle formed by these vectors is half of the area of the parallelogram. Thus

$$A = \frac{1}{2}||(v_{2} - v_{1})\times (v_{3}-v_{1})||$$ makes more sense.

Answer 3

Reverse Engineering

Okay this is what I have until now,

$A = \frac{1}{2}||(v2−v1)X(v3−v1)||$

now vertices can be considered as displacement of point from origin. source: https://stackoverflow.com/a/10640261/8526133

so let us say v1 is at the origin and thus a zero vector for simplification

$$ v2 - v1 = v2 $$ $$ v3 - v1 = v3 $$

so now the equation is

$$\frac{1}{2}||v2 X v3||$$ $$\frac{1}{2} * ||v2|| * ||v3|| * sin\alpha$$

so $||v2||$ becomes the base and $||v3|| sin \alpha $ becomes height

as we know $1/2 * b* h$ and also this is half the area of parallelogram and also the half of the cross product of two vectors.

pardone my drawing skill but here's a supporting image

thanks to @Mohammad Riazi-Kermani, @dxiv, @David C. Ullrich(triangle and parallelogram are not arbitrary).