Proof:

We need to prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular.

Part 1 (Forward Direction):

Let $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$

Expanding the left side:

$$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$$ $$= |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \quad (\text{since } \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a})$$

Given:

$$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$$

Therefore, $2(\vec{a} \cdot \vec{b}) = 0$, which gives $\vec{a} \cdot \vec{b} = 0$

From the property that $\vec{a} \cdot \vec{b} = 0$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular, we conclude that $\vec{a} \perp \vec{b}$.

Part 2 (Reverse Direction):

Let $\vec{a}$ and $\vec{b}$ be perpendicular vectors.

Then $\vec{a} \cdot \vec{b} = 0$ (since perpendicular vectors have dot product zero).

Now,

$$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$$ $$= |\vec{a}|^2 + 2(0) + |\vec{b}|^2$$ $$= |\vec{a}|^2 + |\vec{b}|^2$$

∴ Hence proved.