China Western Mathematical Olympiad

Proof As shown in the figure, we join $E{O}_1$ and $E{O}_2$, then $\angle AE{O}_1=90^{\circ }-\angle ADE$, $\angle BE{O}_2=90^{\circ }-\angle BCE$. Hence $\angle O_{1}E{O}_2=\angle ADE+\angle ECB$. Since $AD\parallel BC$, through $E$ constructing a line parallel to $AD$, we can prove $\angle DEC=\angle ADE+\angle BCE$, so $\angle O_{1}E{O}_2=\angle DEC$. Further, by the sine rule, we can show $$\frac{DE}{EC}=\frac{2O_{1}E\sin A}{2O_{2}E\sin B}=\frac{O_{1}E}{O_{2}E}$$ Thus $△DEC\sim △O_{1}E{O}_2$. Therefore $$\frac{O_1{O}_2}{DC}=\frac{O_{1}E}{DE}=\frac{O_{1}E}{2O_{1}E\sin A}=\frac{1}{2\sin A}$$ So $O_1{O}_2=\frac{DC}{2\sin A}$, which is a fixed value. The proposition is proven.