1. In $\odot O$, there are two inscribed angles $\alpha_{1}, \alpha_{2}$, and the lengths of the chords they intercept are $a_{1}, a_{2}$, respectively. Then, the true proposition among the following is ( , ).
(A) If $\alpha_{1}>\alpha_{2}$, then $a_{1}>a_{2}$
(B) If $\alpha_{1}=\alpha_{2}$, then $a_{1}=a_{2}$
(C) If $a_{1}>a_{2}$, then $\alpha_{1}>\alpha_{2}$
(D) If $a_{1}=a_{2}$, then $\alpha_{1}=\alpha_{2}$
- 1. B.

In figure $a, \alpha_{1}>90^{\circ}>\alpha_{2}$, but $a_{1}<a_{2}$, which can negate (A) and (C). In figure $b, a_{1}=a_{2}$, but $\alpha_{1}<90^{\circ}<\alpha_{2}$, which can negate (D). Therefore, the answer is (B). In fact, in the same circle, equal inscribed angles intercept equal arcs, and equal arcs correspond to equal chords. Hence, the answer is (B).