Arc and Chord Properties:

Theorem 14: In equal circles, if two arcs subtends equal angles at the center, then the arcs are equal. Given: Two circles $C_1 \ and \ C_2$ with centers $O \ and \ P$ respectively.

Two $\widehat{AMB} \ and \ \widehat{CND}$ subtend equal angles $\angle AOB \ and\ \angle CPD$ at the respective centers.

To Prove: $\widehat{AMB} = \widehat{CND}$

Proof:  Join $AB \ and \ CD$

Now consider $\triangle AOB \ and\ \triangle CPD$ $OA = OB$ (radius of the same circle) $CP = PD$ (radius of the same circle)

Now since the two circles are equal, it means that their radius are equal.

Hence $OA = PC \ and \ OB = PD$

Also given that $\angle AOB = \angle CPD$

Hence $\triangle AOB \cong \triangle CPD$ (S.A.S Postulate)

Therefore Now, if you place the two circles on the top of each other, you will see that they coincide and hence $\widehat{AMB} = \widehat{CND}$.

Hence Proved. Note: In the same triangle if $\widehat{AMB} \ and \ \widehat{CND}$ subtend equal angles at the center, then also the arcs are equal. For example: $\angle AOB = \angle COD \Rightarrow \widehat{AMB} =\widehat{CND}$

Theorem 15: In equal circles, if the two arcs are equal, they would subtend equal angles at the center. Given: Two circles $C_1 \ and \ C_2$ with centers $O \ and \ P$ respectively. $\widehat{AMB} = \widehat{CND}$ subtend $\angle AOB \ and\ \angle CPD$ at the respective centers.

To Prove: $\angle AOB = \angle CPD$

Proof: $OA = PC$ (radius of equal circles) $\widehat{AMB} = \widehat{CND}$

Therefore, if you were to place the circles $C_1 \ and \ C_2$ over each other, they will coincide. This also makes that $\angle AOB = \angle CPD$.

Hence Proved.

Theorem 16: In equal circles, if two chords are equal, they will cut equal arcs. Given: Two circles $C_1 \ and \ C_2$ with centers $O \ and \ P$ respectively. Also $Chord \ AB = Chord \ CD$

To Prove: $\widehat{AMB}=\widehat{CND}$

Proof: Consider $\triangle OAB \ and \ \triangle PCD$ $OA = PC$ (radius of equal circles) $OB=PD$ (radius of equal circles) $AB = CD$ (Given)

Therefore $\triangle OAB \cong \triangle PCD$ (S.S.S postulate) $\Rightarrow \angle AOB = \angle CPD$

Therefore $\widehat{AMB}=\widehat{CND}$ ( In equal circles, if two arcs subtends equal angles at the center, then the arcs are equal.)

Hence Proved.

Theorem 17: In two equal circles, if the two arcs are equal the chords of the arcs are also equal. (Converse of Theorem 16) Given: Two circles $C_1 \ and \ C_2$ with centers $O \ and \ P$ respectively. Also $\widehat{AMB}=\widehat{CND}$

To Prove: $Chord \ AB = Chord \ CD$

Proof: Since $\widehat{AMB}=\widehat{CND}$ $\angle AOB = \angle CPD$ (In equal circles, if the two arcs are equal, they would subtend equal angles at the center)

Consider $\triangle AOB \ and \ \triangle CPD$ $\angle AOB = \angle CPD$ $OA=OP$ (radius of equal circles) $OB=PD$ (radius of equal circles) $\Rightarrow \triangle AOB \cong \triangle CPD$ $\Rightarrow Chord \ AB = Chord \ CD$

Hence Proved.