Fig. 50
According to the condition, single-colored diagonals don't have common points. That's why without loss of generality let's consider that the following diagonals are colored in blue: $B_1A_2, B_2A_3, \ldots, B_{n-1}A_n, B_nA_1$, the others are colored in yellow. As $\overrightarrow{B_1A_2} = \vec{h} + \overrightarrow{A_1A_2}$, so $(\overrightarrow{B_1A_2})^2 = (\vec{h})^2 + (\overrightarrow{A_1A_2})^2 + 2 \cdot \vec{h} \cdot \overrightarrow{A_1A_2}$. Similarly, $(\overrightarrow{B_2A_3})^2 = (\vec{h})^2 + (\overrightarrow{A_2A_3})^2 + 2 \cdot \vec{h} \cdot \overrightarrow{A_2A_3}, \ldots, (\overrightarrow{B_{n-1}A_n})^2 = (\vec{h})^2 + (\overrightarrow{A_{n-1}A_n})^2 + 2 \cdot \vec{h} \cdot \overrightarrow{A_{n-1}A_n}$. Let's sum up our equalities:

$$\begin{aligned} B_1 A_2^2 + \dots + B_n A_1^2 &= (\overrightarrow{B_1 A_2})^2 + \dots + (\overrightarrow{B_n A_1})^2 = \\ &= n(\vec{h})^2 + (\overrightarrow{A_1 A_2})^2 + \dots + (\overrightarrow{A_n A_1})^2 + 2 \cdot \vec{h} \cdot (\overrightarrow{A_1 A_2} + \dots + \overrightarrow{A_n A_1}) = nh^2 + A_1 A_2^2 + \dots + A_n A_1^2 \end{aligned}$$
If you make similar calculations for yellow diagonals, then you'll get the same expression.
What was to be demonstrated.