5 Put $\cos \alpha_j$ factors in numerator and denominator where they belong, rather than vice versa, correcting error noted by M.Hardy

I don't know of a "reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle". I do, however, find a proof without explicit induction that to some extent explains the formula, and may be simpler than the proof of original proposer (OP), so it might be of some interest to some readers and perhaps also the OP. It is not simple enough to fit in a comment so I must post it as an answer.

By homogeneity we may assume the circle has diameter 1, which will simplify the formulas. I'll use $j$ rather than $i$ for the index because I'll need $i$ to be $\sqrt{-1}$.

Each $a_j$ is the side of a right triangle with hypotenuse 1 and opposite angle $\alpha_j$. Hence $a_j = \sin \alpha_j$, and the left-hand side is $\sum_{j=1}^n a_j^2$. Each term in the sum on the right-hand side, call it $S$, is $-1/\prod_{j=1}^n -\prod_{j=1}^n \cos \alpha_j$ times $(-1)^{k/2} k$ times a product of $k$ factors $\alpha_j / \cos \alpha_j = \tan \alpha_j$ with distinct $j$'s, for some even integer $k>0$. We might as well include $k=0$ because then $(-1)^{k/2} k = 0$.

To access this sum, consider the finite generating function (or generating polynomial) $$P(X) := \prod_{i=1}^n (1 + i \tan \alpha_j \cdot X).$$ Expand, sum the coefficients, and take the real part. That almost matches $-S$, except that the $X^k$ coefficients are missing the factor $k$. To get that factor, differentiate and set $X=1$: $$-S = \frac1{\prod_{j=1}^n left(\prod_{j=1}^n \cos \alpha_j} alpha_j \right) {\rm Re}\left( P\phantom{|}'(X)|_{X=1}\right) =\frac1{\prod_{j=1}^n \left(\prod_{j=1}^n \cos \alpha_j} alpha_j \right) {\rm Re}\left( P(1) \phantom{/}\sum_{j=1}^n \frac{i \tan \alpha_j}{1 + i \tan \alpha_j \cdot X} \Biggl|_{X=1}\Biggr.\right).$$ But $P(1)$ is the product of terms $1/(1+i\tan\alpha_j) 1+i\tan\alpha_j = e^{i \alpha_j} / \cos \alpha_j$, and (aha) $\sum_j \alpha_j = \pi$ because each $\alpha_j$ is half the angle subtended by the $j$-th side about the center of the circle. Hence $P(1)$ is the real number $- -1 / \prod_{j=1}^n \cos \alpha_j$, and $S$ simplifies to $$+ \sum_{j=1}^n \phantom| {\rm Re} \frac{i \tan \alpha_j}{1 + i \tan \alpha_j} = \sum_{j=1}^n \phantom| \frac{\tan^2 \alpha_j} {1+\tan^2 \alpha_j} = \sum_{j=1}^n \phantom| \sin^2 \alpha_j,$$ QED.

4 inserted \cdot in $i \tan \alpha_j \cdot X$ (twice)

I don't know of a "reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle". I do, however, find a proof without explicit induction that to some extent explains the formula, and may be simpler than the proof of original proposer (OP), so it might be of some interest to some readers and perhaps also the OP. It is not simple enough to fit in a comment so I must post it as an answer.

By homogeneity we may assume the circle has diameter 1, which will simplify the formulas. I'll use $j$ rather than $i$ for the index because I'll need $i$ to be $\sqrt{-1}$.

Each $a_j$ is the side of a right triangle with hypotenuse 1 and opposite angle $\alpha_j$. Hence $a_j = \sin \alpha_j$, and the left-hand side is $\sum_{j=1}^n a_j^2$. Each term in the sum on the right-hand side, call it $S$, is $-1/\prod_{j=1}^n \cos \alpha_j$ times $(-1)^{k/2} k$ times a product of $k$ factors $\alpha_j / \cos \alpha_j = \tan \alpha_j$ with distinct $j$'s, for some even integer $k>0$. We might as well include $k=0$ because then $(-1)^{k/2} k = 0$.

To access this sum, consider the finite generating function (or generating polynomial) $$P(X) := \prod_{i=1}^n (1 + i \tan \alpha_j \cdot X).$$ Expand, sum the coefficients, and take the real part. That almost matches $-S$, except that the $X^k$ coefficients are missing the factor $k$. To get that factor, differentiate and set $X=1$: $$-S = \frac1{\prod_{j=1}^n \cos \alpha_j} {\rm Re}\left( P\phantom{|}'(X)|_{X=1}\right) = \frac1{\prod_{j=1}^n \cos \alpha_j} {\rm Re}\left( P(1) \phantom{/}\sum_{j=1}^n \frac{i \tan \alpha_j}{1 + i \tan \alpha_j \cdot X} \Biggl|_{X=1}\Biggr.\right).$$ But $P(1)$ is the product of terms $1/(1+i\tan\alpha_j) = e^{i \alpha_j} \cos \alpha_j$, and (aha) $\sum_j \alpha_j = \pi$ because each $\alpha_j$ is half the angle subtended by the $j$-th side about the center of the circle. Hence $P(1)$ is the real number $- \prod_{j=1}^n \cos \alpha_j$, and $S$ simplifies to $$+ \sum_{j=1}^n \phantom| {\rm Re} \frac{i \tan \alpha_j}{1 + i \tan \alpha_j} = \sum_{j=1}^n \phantom| \frac{\tan^2 \alpha_j} {1+\tan^2 \alpha_j} = \sum_{j=1}^n \phantom| \sin^2 \alpha_j,$$ QED.

3 Corrected a sign error in the initial formula for $S$

I don't know of a "reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle". I do, however, find a proof without explicit induction that to some extent explains the formula, and may be simpler than the proof of original proposer (OP), so it might be of some interest to some readers and perhaps also the OP. It is not simple enough to fit in a comment so I must post it as an answer.

By homogeneity we may assume the circle has diameter 1, which will simplify the formulas. I'll use $j$ rather than $i$ for the index because I'll need $i$ to be $\sqrt{-1}$.

Each $a_j$ is the side of a right triangle with hypotenuse 1 and opposite angle $\alpha_j$. Hence $a_j = \sin \alpha_j$, and the left-hand side is $\sum_{j=1}^n a_j^2$. Each term in the sum on the right-hand side, call it $S$, is $1/\prod_{j=1}^n -1/\prod_{j=1}^n \cos \alpha_j$ times $(-1)^{k/2} k$ times a product of $k$ factors $\alpha_j / \cos \alpha_j = \tan \alpha_j$ with distinct $j$'s, for some even integer $k>0$. We might as well include $k=0$ because then $(-1)^{k/2} k = 0$.

To access this sum, consider the finite generating function (or generating polynomial) $$P(X) := \prod_{i=1}^n (1 + i \tan \alpha_j X).$$ Expand, sum the coefficients, and take the real part. That almost matches $-S$, except that the $X^k$ coefficients are missing the factor $k$. To get that factor, differentiate and set $X=1$: $$-S = \frac1{\prod_{j=1}^n \cos \alpha_j} {\rm Re}\left( P\phantom{|}'(X)|_{X=1}\right) = \frac1{\prod_{j=1}^n \cos \alpha_j} {\rm Re}\left( P(1) \phantom{/}\sum_{j=1}^n \frac{i \tan \alpha_j}{1 + i \tan \alpha_j X} \Biggl|_{X=1}\Biggr.\right).$$ But $P(1)$ is the product of terms $1/(1+i\tan\alpha_j) = e^{i \alpha_j} \cos \alpha_j$, and (aha) $\sum_j \alpha_j = \pi$ because each $\alpha_j$ is half the angle subtended by the $j$-th side about the center of the circle. Hence $P(1)$ is the real number $- \prod_{j=1}^n \cos \alpha_j$, and $S$ simplifies to $$+ \sum_{j=1}^n \phantom| {\rm Re} \frac{i \tan \alpha_j}{1 + i \tan \alpha_j} = \sum_{j=1}^n \phantom| \frac{\tan^2 \alpha_j} {1+\tan^2 \alpha_j} = \sum_{j=1}^n \phantom| \sin^2 \alpha_j,$$ QED.

2 A bit more copy editing, mostly to highlight and explain the key point $\sum_j \alpha_j = \pi$
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