AoPS - Problem

Indeed its an Elementary Geo Solution : Let $\angle{ABE}=\alpha, \angle{EBD}=\beta , BE \cap AC = X$. Now since $AB$ is tangent to $\omega_2$ so $$\angle{BO_2E}=2.\angle{BDE}=2.\angle{ABE}=2\alpha$$Again we know , $$\angle{DO_2E}=2.\angle{DBE}=2\beta \Longrightarrow \angle{BO_2D}=\angle{DO_2E}+\angle{BO_2E}=2\alpha+2\beta$$Notice that $$O_1A \perp AB , O_2B \perp AB \Longrightarrow O_1A || O_2B \Longrightarrow \angle{AO_1D}=180-\angle{BO_2D}=180-2(\alpha+\beta)$$Now observe that using PoP we get $$O_1D.O_2D=DE.DC \Longrightarrow \triangle {DO_1C} \sim \triangle{DO_2E} $$$$\Longrightarrow \angle{DO_1C}=\angle{DO_2B}=2\beta \Longrightarrow \angle{AO_1C}=\angle{DO_1C}+\angle{AO_1D}=180-2\alpha $$Since $$AO_1=CO_1 \Longrightarrow \angle{O_1AC}=\alpha \Longrightarrow \angle{BAX}=\angle{BAO_1}-\angle{XAO_1}=90-\alpha \Longrightarrow \angle{AXB}=90$$So we are done !