Hope this makes sense. Let the tangent points of the incircle of ABC and on the sides BC, AC, and AB be D, E, and F. Also let the intersection of the incircle of ABC and BI to be X. Now since DI=DA', IE=EB', ID=DC', and IDA', IEB', and IDC' are straight lines all by the definition of a reflection, then A'B' // DE, A'C' // DF, and B'C' // EF. In other words, DEF is similar to A'B'C', and is in the ratio 1:2. So, a dilation taking D to A', E to B', and F to C' will also take the circumcircle of DEF to the circumcircle of A'B'C', taking X to B. So, looking at triangle ID, we have DI=r (r is the radius of the circumcircle of DEF) and BI=2r, with <BDI=90. So, we have a 30-60-90 triangle, with <IBD=30. Similarly, <FBI=30, so angle ABC=60

isnt the solution just angle chasing? kennyworm

there is another solution. it very palpable that IA'=IB'=IC' <C'IB'=180-<A <A'IB'=180-<C <A'BC'=2<B so <A'B'C'=1/2 (<A+<C) since A',B',C', B are on the same circle <C'BA'+<A'B'C'=180 2<B+1/2(<A+<C)=180 3/2<B+1/2(<A+<B+<C)=180 3/2<B=90 <B=60

Why not simply: I is the circumcentre of A'B'C' (Erratum: was ABC). Thus IB = 2r. Thus IBC = 30deg... and so ABC = 60 deg.

Yes , you are right kueh.It is so simple. By the way you have typo in your post : I is the circumcentre of A'B'C'.

ah yes, thanks, i changed it now.