Sums of vector triangles and a resultant identity of determinants

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In the $\mathbb R^2$ plane, let $a, b, c$ be vectors, let $o$ be the origin, and let $d = b + c$. Determine the area of $\triangle oad$ in terms of $\triangle oab$ and $\triangle oac$, and infer an identity regarding the determinant of a sum. (Source: Inspired by this problem.)

My solution is below. I request verification, critique, or alternate approaches.

WLOG, assume $a = i$ (that is, $[1 \quad 0]^\top$). Since the area of triangles on the same base depends only on height, the area of $\triangle oab$ is a directly proportional function of $b_2$ (but independent of $b_1$). Similarly, the area of $\triangle oac$ is a directly proportional function of $c_2$.

Consider the case when $b$ and $c$ are in the same half-plane induced by $a$ (that is, either both above or both below). Then the height of $\triangle oad$ equals $b_2 + c_2$, and so the area of $\triangle oad$ equals the sum of the areas of $\triangle oab$ and $\triangle oac$.

In the other case, with $b$ and $c$ in different half-planes, the height of $\triangle oad$ equals $b_2 - c_2$, and the area of $\triangle oad$ is the difference in areas of $\triangle oab$ and $\triangle oac$.

We can simplify the above by using signed areas. Then, in either case, the signed area of $\triangle oad$ equals the sum of the signed areas of $\triangle oab$ and $\triangle oac$.

Since determinants in $\mathbb R^2$ are equal to signed area, this shows that $$\det([A \quad (B+C)]) = \det([A \quad B]) + \det([A \quad C])$$ that is, determinants are linear in each column.

I'm not able to determine if this identity generalizes.

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Since you're working in a vector space already it follows that the area of a triangle composed of vectors $\triangle 0 \vec a \vec d = \frac{1}{2} | \vec a \times \vec d |$.

Then cross product identities give

$$ \frac{1}{2} | \vec a \times \vec d | = \frac{1}{2} | \vec a \times (\vec b + \vec c) | $$

$$ = \frac{1}{2} | \vec a \times \vec b + \vec a \times \vec c | $$

$$ \le \frac{1}{2} | \vec a \times \vec b | + \frac{1}{2} | \vec a \times \vec c|$$

Finally the determinant of a matrix is related to the cross product in a natural way.