This problem was proposed by Burii.

it is enough if we prove that the periferal angles corresponding a + the one corresponding c = the one corresponding b + the one correspondin d Proof: just remind that arc = (perimeter of circle) * (central angle corresponding that arc) / 360 (x) = periferic angle on arc x. (a) =( <AK1N2 - <K1N2K2 + <AN2K1 - <N2K1N1) / 2 = (<C - (K1K2) - (N1N2)) /2 similarly (c) = (<A - (L2L1) - (M1M2))/2 (a) + (c) = (180 - (K1K2) - (N1N2) - (L2L1) - (M1M2)) / 2 similarly (b) + (d) = (180 - (K1K2) - (N1N2) - (L2L1) - (M1M2)) / 2 = (a) + (c) DOne

https://om.mimuw.edu.pl/static/app_main/problems/om70_2r.pdf The official solution

ryan17 wrote: A cyclic quadrilateral $ABCD$ is given. Point $K_1, K_2$ lie on the segment $AB$, points $L_1, L_2$ on the segment $BC$, points $M_1, M_2$ on the segment $CD$ and points $N_1, N_2$ on the segment $DA$. Moreover, points $K_1, K_2, L_1, L_2, M_1, M_2, N_1, N_2$ lie on a circle $\omega$ in that order. Denote by $a, b, c, d$ the lengths of the arcs $N_2K_1, K_2L_1, L_2M_1, M _2N_1$ of the circle $\omega$ not containing points $K_2, L_2, M_2, N_2$, respectively. Prove that \begin{align*} a+c=b+d. \end{align*}

Let $h$ be homothety with positive scale that maps circle $(ABCD)$ to $\omega$. Let $A'=h(A), B'=h(B), C'=h(C), D'=h(D)$. Then $AB \parallel A'B'$, hence $\mathrm{arc}\ A'K_1=\mathrm{arc}\ K_2B'$. Similarly $\mathrm{arc}\ B'L_1=\mathrm{arc}\ L_2C'$, $\mathrm{arc}\ C'M_1=\mathrm{arc}\ M_2D'$, and $\mathrm{arc}\ D'N_1=\mathrm{arc}\ N_2A'$. Hence $a+c=b+d$. 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