In triangle ABC the angle B is 60° and the sides satisfy $b^2$ = ac. The triangle must be:

The cosine rule at $B$ gives $b^2=a^2+c^2-2ac$·cos60° = $a^2+c^2-ac$. Setting that equal to the given $ac$ leaves $a^2+c^2-2ac=0$, i.e. $(a-c{)}^2=0$, so $a=c$. A triangle with $a=c$ has $A=C$, and with $B=60$° the remaining 120° splits evenly, making all three angles 60° — the triangle is equilateral, and then $b^2=ac$ holds trivially. The condition therefore does not merely force 'isosceles': the fixed 60° at $B$ upgrades it all the way. A 60° angle plus two equal sides can never leave an obtuse or a non-isosceles right-angled triangle.