Part(a)
Let ABC is given triangle . we draw equilateral triangles ARB , BPC and CQA on the sides AB , BC and CA of ABC.
Let us draw circumcircles of ARB and BPC . Let it intersect at T .
Quadrilateral ARBT and Quadrilateral BPCT are cyclic .
Hence ATB = BTC = 120 because ARB = BPC = 60 .
Hence ATC = 120 and quadrilateral AQCT is cyclic because AQC = 60 .
Hence the Toricelli point is intersection of circumcircles of ARB , BPC and CQA .
------------------------------------------------------
To prove AP , BQ and CR contains T means AP , BQ and CR are straight lines and T is common intersection point.
It is already explained that ATC = 120 . We see from the figure PBC and PTC are angle subtended by chord PC in a circle . since PBC = 60 , then PTC = 60 . Hence ATC + PTC = 180 and AP is a straight line.
Similarly we can prove BQ and CR are straightlines . Hence T is common intersection point of AP, BQ and CR
question is based from book: Euclidean Plane and its Relatives, A Minimalist Introduction by Anton Petrunin...