we have $a^2+b^2+c^2= ab + bc + ca$
Hence $a^2+b^2+c^2 -ab - bc -ca = 0$
Hence $2a62+2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0$
Or $(a^2-2ab + b^2) + (b^2-2bc+c^2) + (c^2-2ac+a^2) = 0$
or $(a-b)^2 + (b-c)^ + (c-a)^2 = 0$
as each term is non -ve so each of then has to be zeo so a= b= c and hence the triange is equilateral