In a circle, chord AB is congruent to chord CD. They intersect at point S. Prove that arc AC is congruent to arc BD.

Question

I presume that one should show the angle B is congruent to the angle A. However, in the figure I made, the angle A is congruent to the angle C.

Answer 1

Let $A\in\smallsmile CD$.

Thus, $D\in\smallsmile AB$. (It means $ADBC$ is a quadrilateral).

Id est, $$\widehat{AC}=\widehat{CD}-\widehat{AD}=\widehat{AB}-\widehat{AD}=\widehat{BD}.$$

Answer 2

Two congruent chords $AB$ and $CD$ in a circle are equal in distance from the center. These chords interesct at $S$, and have midpoints $E$ and $F$ respectively. The perpendicular bisector of a chord in a circle goes through the origin, or equivalently a radius perpendicular to a chord bisects the chord. Here $EO$ and $FO$ are perpendicular to $AB$ and $CD$ respectively, and are equal since $AB$ and $CD$ are congruent, and so an equal distance from the centre $O$.

$AB\cong CD$

$OA\cong OB\cong OC\cong OD$

Hence triangles $AOB$ and $COD$ are similar, and so $\angle AOB \cong \angle COD$. Hence $\overparen{AB}\cong\overparen{CD}$ since the arcs subtend equivalent central angles.

Now to prove the claim:

Proof.1

We note:

$\angle COD =\angle AOB$

$\angle AOD = \angle AOB-\angle DOB$

$\angle AOD = \angle COD-\angle AOC$

Hence $\angle AOB-\angle DOB=\angle COD-\angle AOC$, but since $\angle AOB \cong \angle COD$ this becomes

$\angle AOC=\angle BOD$

Hence $\overparen{AC}\cong\overparen{BD}$ since the arcs subtend the equivalent central angles $\angle AOC$ and $\angle BOD$ respectively.

Proof.2

We have $\overparen{AB}\cong\overparen{CD}$. Hence

$\overparen{AD}:\overparen{AC}\propto\overparen{AD}:\overparen{DB}\implies\overparen{AC}\cong\overparen{DB}$