Two circles having their centres at O and P intersect each other at Q and R. Two straight lines AQB and CRD are drawn parallel to OP. Prove that 

(i) AB=2OP

(ii) AB=CD

 

Construction: Join Ol is perpendicular to AB and Pl' is also perpendicular to AB.

Proof: As Ol⊥AB

⇒Ol⊥AQ  

⇒Al=lQ

⇒AQ=2lQ  .........(i)

Again,

 Ol'⊥AB

⇒Ol'⊥QB

⇒Ql'=l' B

⇒QB=2Ql'  ........(ii)

⇒AB=AQ+QB=2lQ+2Ql'  [From (i) & (ii)]

⇒AB=2(lQ+Ql')

⇒AB=2l l'

⇒AB=2OP  ..........(I)

Hence the result.

Similarly, draw OM ⊥CD and PM' ⊥CD.

Again ,

OM⊥CD

⇒OM⊥ CR

⇒CM=MR

⇒CR=2MR............(iii)

Now, PM' ⊥CD

⇒PM' ⊥RD

⇒RM'=M'D

⇒RD=2RM'............(iv) 

CD=CR+RD=2(MR+RM')  [From (iii) & (iv)]

CD=MM'

CD= 2OP ........(II)

From (I) and (II), we get,

AB= CD.

Hence the result.

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