Arc and Chord Properties:

Theorem 14: In equal circles, if two arcs subtends equal angles at the center, then the arcs are equal.c35

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.

c34

 

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.c33

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.c32

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)c31

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.

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