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In an acute-angled triangle $\mathrm{ABC}$, the altitudes from $\mathrm{A, B, C}$ when extended intersect the circumcircle again at points $A_{1}, B_{1}, C_{1}$, respectively. If $\angle A B C=45^{\circ}$ then $\angle A_{1}, B_{1}, C_{1}$ equals
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$90^{\circ}$
$\angle \mathrm{BCH}=45^{\circ}=\angle \mathrm{BCA}_{1}$
$\angle \mathrm{C}_{1} \mathrm{CA}_{1}=\angle \mathrm{C}_{1} \mathrm{B}_{1} \mathrm{A}_{1}=90^{\circ}$
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