Three equal circles thouch one another extenally. Prove that the centres of the three circles form an equilateral triangle.

Explanation

Let three equal circles with centres A, B and C touch one another extentally at the points P, Q and R.
To prove : `DeltaABC` is an equilateral tringle.
Proof In the circle with centre A,
`PA = AR[because` radii of same circle]
Similarly, `BP =BQ=CR.`
Again, the circles are equal, so the radii of them are al equa, i.e., `AP=BP=BQ=CQ=CR=RA......(1)`
Now, `AB=AP +BP=BQ+CQ=BC......(2)`
`[because AP=BQand BP=CQ,` from (1)]
Again, `BC=BQ+CQ=CR+AR[becauseBQ=CRand CQ=AR]=CA.........(3)`
`therefore ` from (2) and (3) we get, `AB=BC=CA.`
`therefore DeltaABC` is an equilateral triangle. (Proved)