Midpoints of arcs in cyclic quadrilateral

Question

Do you have any hint? I don't want a solution, please. Only some hints.

Let $A,B,C,D$ the four vertex of a cyclic quadrilateral, and $E,H,F,G$ the midpoints of arcs $AB$, $BC$, $CD$ and $DA$. Show that $EF\perp GH$

Answer 1

Since the angle subtended by an arc at the circle is proportional to the arc, for arcs $\overset{ \huge\frown}{ABC} = 2\overset{ \huge\frown}{EBH}$,

$$\angle ADC = 2\angle EFH$$

And for arcs $\overset{ \huge\frown}{ADC} = 2\overset{ \huge\frown}{GDF}$,

$$\angle ABC = 2\angle GHF$$

Consider the cyclic quadrilateral $ABCD$, its opposite angles add to $180^\circ$,

$$\begin{align*} \angle ADC + \angle ABC &= 180^\circ\\ \angle EFH + \angle GHF &= 90^\circ \end{align*}$$

So $\triangle FIH$ is a right-angled triangle, where $I$ is the intersection of $EF$ and $GH$.