IRL_ABooklet

Connect $C$ to $F$ and $B$ to $G$. This produces three cyclic quadrilaterals $ABGH$, $BCFG$, and $CDEF$. Because opposite angles in a cyclic quadrilateral add to $180^{\circ }$, we see that $\angle A+\angle C+\angle E+\angle G=3\cdot 180^{\circ }$ which is half of the interior angle sum of an octagon. Therefore, $\angle A+\angle C+\angle E+\angle G=\angle B+\angle D+\angle F+\angle H$.

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Alternative solution.

Let $O$ be the centre of the circle and join $O$ to the vertices of the octagon. This way we obtain eight isosceles triangles. The connection from $O$ to a vertex splits the internal angles of the octagon at this vertex into two angles. For example, $\angle BAH=\angle BAO+\angle OAH$. These two angles are base angles of two different isosceles triangles. The other base angle of each of these isosceles triangles contributes to the internal angle of a neighbouring vertex. For example, $\angle OBA$ contributes to $\angle B$ and $\angle OHA$ contributes to $\angle H$. Therefore, the sum of the interior angles at $A,C,E,G$ is the same as the sum of every second interior angle starting at $H$, i.e. $\angle A+\angle C+\angle E+\angle G=\angle B+\angle D+\angle F+\angle H$.