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, $\angle A + \angle C + \angle E + \angle G = \angle B + \angle D + \angle F + \angle H$.

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. $\angle A + \angle C + \angle E + \angle G = \angle B + \angle D + \angle F + \angle H$.