Question

In the given figure, common tangents AB and CD to the two circles with centres O₁ and O₂ intersect at E. Prove that : AB = CD.

Answer 1

Let C₁ & C₂ are circles whose centers are O₁ and O₂ respectively. 

And AB and CD are tangents to both circles C₁ & C₂. 

We know that lengths of tangents from an external point to circle are same. 

∴ EA = EC. … (1)    (EA and EC are tangents to circle C₁ form external point E.) 

And EB = ED. … (2)    (EB and ED are tangents to circle C₂ from external point E.) 

Now, adding equation (1) & (2), we get EA + EB = EC + ED 

⇒ AB = CD. (∵ AB = AE + EB, AE = EA & CD = CE + ED, CE = EC) 

Hence proved.