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.