The Problems of Ukrainian Authors

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: $B_1{A}_2,B_2{A}_3,\dots ,B_{n-1}{A}_n,B_n{A}_1$, the others are colored in yellow. As $\overrightarrow{B_1{A}_2}=\vec{h}+\overrightarrow{A_1{A}_2}$, so $(\overrightarrow{B_1{A}_2}{)}^2=(\vec{h}{)}^2+(\overrightarrow{A_1{A}_2}{)}^2+2\cdot \vec{h}\cdot \overrightarrow{A_1{A}_2}$. Similarly, $(\overrightarrow{B_2{A}_3}{)}^2=(\vec{h}{)}^2+(\overrightarrow{A_2{A}_3}{)}^2+2\cdot \vec{h}\cdot \overrightarrow{A_2{A}_3},\dots ,(\overrightarrow{B_{n-1}{A}_n}{)}^2=(\vec{h}{)}^2+(\overrightarrow{A_{n-1}{A}_n}{)}^2+2\cdot \vec{h}\cdot \overrightarrow{A_{n-1}{A}_n}$. Let's sum up our equalities:

$$\begin{array}{rl}{B}_1{A}_2^2+\cdots +B_n{A}_1^2& =(\overrightarrow{B_1{A}_2}{)}^2+\cdots +(\overrightarrow{B_n{A}_1}{)}^2=\\ & =n(\vec{h}{)}^2+(\overrightarrow{A_1{A}_2}{)}^2+\cdots +(\overrightarrow{A_n{A}_1}{)}^2+2\cdot \vec{h}\cdot (\overrightarrow{A_1{A}_2}+\cdots +\overrightarrow{A_n{A}_1})=n{h}^2+A_1{A}_2^2+\cdots +A_n{A}_1^2\end{array}$$ If you make similar calculations for yellow diagonals, then you'll get the same expression. What was to be demonstrated.