$A^{2 .} \triangle ABC$ 's three interior angles $A, B, C$ form an arithmetic sequence, and the heights $h_{a}, h_{b}, h_{c}$ on the three sides $a, b, c$ also form an arithmetic sequence. Then $\triangle ABC$ is ( ).
(A) Isosceles but not equilateral triangle
(B) Equilateral triangle
(C) Right triangle
(D) Obtuse non-isosceles triangle
2. B. From $A, B, C$ forming an arithmetic sequence, we know that $B=60^{\circ}$. Let the area of $\triangle A B C$ be $S$. Then, from $2 h_{b}=h_{a}+h_{c}$, we get $\frac{2 S}{b}=\frac{S}{a}+\frac{S}{c}$. Therefore, $2 a c=a b+b c$.

By the Law of Sines, and noting $A=60^{\circ}-\theta, C=60^{\circ}+\theta, 0^{\circ} \leqslant \theta \leq 0$, solving yields $\cos \theta=1, \theta=0^{\circ}$.
Thus, $A=C=60^{\circ}$.
Hence, $\triangle A B C$ is an equilateral triangle.