Given matrices A = [[3, 1], [6, 2]] and B = [[-1, 3], [3, 9]], calculate (A+B)A.

Let $C = A+B$. We need to calculate $CA$.

The final answer is $\boxed{\begin{pmatrix} 30 & 10 \\ 93 & 31 \end{pmatrix}}$.

In a parallelogram ABCD, opposite sides are equal and parallel, so $\vec{DC} = \vec{AB}$.
Substitute $\vec{DC}$ with $\vec{AB}$ in the left side of the equation:

H is the midpoint of BC. This means $\vec{BH} = \frac{1}{2}\vec{BC}$.
In a parallelogram ABCD, $\vec{BC} = \vec{AD}$.
So, $\vec{BH} = \frac{1}{2}\vec{AD}$.

Now, express $\vec{AH}$ using the triangle rule for vector addition:

Substitute $\vec{BH} = \frac{1}{2}\vec{AD}$:

From $(*)$ we have $\vec{AB} + \vec{AD} + \vec{DC} = 2\vec{AB} + \vec{AD}$.
From $(**)$ we have $2\vec{AH} = 2\vec{AB} + \vec{AD}$.
Since both sides simplify to the same expression, the equality is proven.
Therefore, $\boxed{\vec{AB} + \vec{AD} + \vec{DC} = 2\vec{AH}}$.