In an acute-angled triangle ABC,the altitudes | vedclass

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(C) Let $H$ be the orthocenter of $	riangle ABC$. The altitudes $AD, BE, CF$ are extended to meet the circumcircle at $A_1, B_1, C_1$.In $	riangle A

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$90$

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(C) Let $H$ be the orthocenter of $	riangle ABC$. The altitudes $AD, BE, CF$ are extended to meet the circumcircle at $A_1, B_1, C_1$.In $	riangle ABD$,$\angle ADB = 90^{\circ}$ and $\angle ABD = 45^{\circ}$,so $\angle BAD = 45^{\circ}$.Since $A, B, A_1, C_1$ lie on the circumcircle,$\angle B A_1 A = \angle B C_1 A = \angle B C A$.Also,$\angle B B_1 A_1 = \angle B A A_1 = \angle BAD = 45^{\circ}$ (angles subtended by the same arc $BA_1$).Similarly,$\angle B B_1 C_1 = \angle B C C_1 = 45^{\circ}$ (angles subtended by the same arc $BC_1$).Therefore,$\angle A_1 B_1 C_1 = \angle B B_1 A_1 + \angle B B_1 C_1 = 45^{\circ} + 45^{\circ} = 90^{\circ}$.

In an acute-angled $\triangle ABC$,the altitudes from $A, B, C$ when extended intersect the circumcircle again at points $A_1, B_1, C_1$ respectively. If $\angle ABC = 45^{\circ}$,then $\angle A_1 B_1 C_1$ equals (in $^{\circ}$)

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