How to prove the circumferential angle is equal if the arcs are equal in equal circles?

Question

How to prove the circumferential angle is equal if the arcs are equal in equal circles?

The $\frown{A}{B}$ equals to $\frown{C}{D}$, how to prove the $\angle{AFB}$ = $\angle{CED}$?

and why if $\angle{AFB}$ = $\angle{CED}$, then we know $\frown{A}{B}$ equals to $\frown{C}{D}$?

Answer 1

There are fife cases for the proof of the theorem about an inscribed angle.

In the case for the your picture we can make the following.

Let $O$ be a center of the circle.

Thus, $$\measuredangle CFD=\measuredangle OED-\measuredangle OEC=\frac{180^{\circ}-\measuredangle EOD}{2}-\frac{180^{\circ}-\measuredangle EOC}{2}=$$ $$=\frac{\measuredangle EOC-\measuredangle EOD}{2}=\frac{1}{2}\measuredangle DOC=\frac{1}{2}\measuredangle AOB=\measuredangle AFB.$$

Answer 2

I will use the pure Euclidean Geometry.

If the arcs $AB$ and $CD$ are congruent, then, called $O$ the center of the circumference, the angles $AOB$ and $COD$ are also congruent. Therefore the circumferential angles are congruent, because they are congruent to half angles which are congruent, namely $AOB$ and $COD$.

Conversely, if the circumferential angles are congruent, then the angles $AOB$ and $COD$ are congruent, because they are congruent to the double of angles which are congruent (the circumference angles). So, since the angles with vertex in $O$ are congruent, also the correspondent arcs $AB$ and $CD$ are congruent. QED