Distributive Property of Dot Products - Geometrically, but a different approach. The Question:
The distributive property for dot-products says that:
$$\vec{a} \cdot (\vec{b} + \vec{c})=\vec{a} \cdot \vec{b} + \vec{a} \cdot\vec{c}$$
If we think of the dot product as the projections of $\vec{b}$ and $\vec{c}$ onto $\vec{a}$ scaled by the magnitude of $\vec{a}$, this makes a lot of geometric sense. 
                                                                      
The amount which $\vec{b}$ moves in the direction of $\vec{a}$ (scaled by the magnitude of $\vec{a}$) plus the amount which $\vec{c}$ moves in the direction of $\vec{a}$ (scaled by the magnitude of $\vec{a}$) is equal to the amount which their sum, $(\vec{b} + \vec{c})$, moves in the direction of $\vec{a}$ (scaled by the magnitude of $\vec{a}$).
Both these answers on SE have used that explanation.
Proving that the dot product is distributive?
Prove the distributive property of the dot product using its geometric definition?
However, what if instead of considering the projection of $(\vec{b} + \vec{c})$ onto $\vec{a}$, we interpret the dot product as the projection of $\vec{a}$ onto the other two vectors, scaled by their magnitudes?
In that case, the distributive property would be saying that the projection of $\vec{a}$ onto $\vec{b}$ scaled by the magnitude of $\vec{b}$ plus the projection of $\vec{a}$ onto $\vec{c}$ scaled by the magnitude of $\vec{c}$ is equal to the projection of $\vec{a}$ onto $(\vec{b}+\vec{c})$ scaled by the magnitude of $(\vec{b}+\vec{c})$.
                                                              
However, after having drawn out a bunch of pictures, I haven't been able to geometrically justify this, and I really would appreciate some help!
Written out more symbolically, while its obvious that...
$$\|(B+C)_A\|*\|A\|=\|B_A\|*\|A\|+|C_A\|*\|A\|$$
...to me, its not so obvious that:
$$\|A_{(B+C)}\|*\|(B+C)\|=\|A_B\|*\|B\|+|A_C\|*\|C\|$$
I hope I'm explaining the question well...please let me know in the comments if I should make edits to make the question clearer.
Thank you.

Edit:
I'll explain why I haven't accepted an answer yet. I understand what the answers are saying. Pretty much, that if we accept the commutative property of dot products, then the desired result follows logically. 
So first, I'll explain the commutative property as intuitively as possible (I typed it out in Microsoft word, so I'm just pasting in screenshots)


Okay. That's the commutative property. I didn't draw out the vectors, but hopefully its easy to picture them as you follow it.
Finally, if I understand the argument the answers have given so far, its this:

That's all good, and I can follow it logically! But...isn't there a way to make "IV" pop intuitively, in a way similar to the way in which the pictures do so for "I"? 
Or, at least an example that makes it really intuitive? I can't come up with one...
Thanks again!
 A: You can write the distributive property in the form
$$\vec a \cdot (\vec b + \vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c$$
but you can alternatively and equally well write it as
$$(\vec b + \vec c) \cdot \vec a = \vec b \cdot \vec a + \vec c \cdot \vec a.$$
These equations are equivalent, of course, because the dot product is commutative.
Geometrically, for $\vec a \cdot (\vec b + \vec c)$ you can project $\vec b$ and $\vec c$ (and their sum) onto $\vec a$, while for $(\vec b + \vec c) \cdot \vec a$ you can again project $\vec b$ and $\vec c$ (and their sum) onto $\vec a$.
That's the implication of commutativity--it is just as valid to project the left-hand vector onto the right-hand one as to project the right-hand vector onto the left-hand one.
Therefore
$$(\vec a+\vec b) \cdot \vec c = \vec a \cdot \vec c + \vec b \cdot \vec c$$
and
$$(\vec a+\vec b) \cdot \vec d = \vec a \cdot \vec d + \vec b \cdot \vec d,$$
via the exact same geometric interpretation as before,
completing your proof.
A: Two non-zero, non-orthogonal vectors $A, B,$ with projections $A_B, B_A,$ determine similar right-angled triangles, therefore:
$$
\frac{\|A\|}{\|A_B\|} = \frac{\|B\|}{\|B_A\|}.
$$
Threfore, for any two vectors $A,B$ (which may even be zero, or orthogonal):
$$
\|A_B\| \cdot \|B\| = \|B_A\| \cdot \|A\|.
$$
Therefore, if $A, B, C$ are any three vectors:
\begin{align*}
\|A_{B + C}\| \cdot \|B + C\| & = \|(B + C)_A\| \cdot \|A\| \\
& = (\|B_A\| + \|C_A\|) \cdot \|A\| \\
& = \|B_A\| \cdot \|A\| + \|C_A\| \cdot \|A\| \\
& = \|A_B\| \cdot \|B\| + \|A_C\| \cdot \|C\|.
\end{align*}
