6. Given an acute-angled triangle $A B C$. The height $A A_{1}$ is extended beyond vertex $A$ to segment $A A_{2}=B C$. The height $C C_{1}$ is extended beyond vertex $C$ to segment $C C_{2}=A B$. Find the angles of triangle $A_{2} B C_{2}$. ( ( . Zhenedarov)
Answer. $\angle A_{2} B C_{2}=\angle 90^{\circ}, \angle B A_{2} C_{2}=\angle B C_{2} A_{2}=45^{\circ}$.

Solution. Triangles $A B A_{2}$ and $C C_{2} B$ are equal by two sides and the angle between them $\left(A B=C C_{2}, A A_{2}=B C\right.$, and angles $B A A_{2}$ and

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$B C C_{2}$ are equal as adjacent to angles $A_{1} A B$ and $C_{1} C B$, which are equal to $90^{\circ}-\angle A B C$). Therefore, $\angle B A_{2} A=\angle C B C_{2}$, from which $\angle A_{2} B C_{2}=\angle A_{2} B A+\angle A B C+\angle C B C_{2}=\angle A B C+\left(\angle A_{2} B A+\angle B A_{2} A\right)=\angle A B C+\angle B A A_{1}=90^{\circ}$. Moreover, $A_{2} B=C_{2} B$, from which $\angle B A_{2} C_{2}=\angle B C_{2} A_{2}=\left(180^{\circ}-\angle A_{2} B C_{2}\right) / 2=45^{\circ}$.

Remark. The statement of the problem remains true for a non-acute triangle $A B C$.