Question

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

• A. ${45}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${135}^o$

Answer 1

The correct option is C ${90}^o$

$In△BAD$

By angle sum property

$\angle ABD+\angle BAD+\angle ADB=18045+\angle BAD+90=180$

$\angle BAD=45or\angle BA{A}^{\prime }=45\angle BA{A}^{\prime }=\angle B{B}^{\prime }{A}^{\prime }=45$

Angles made in the same segment of the circle

Similarly in $△BCF$

$\angle FBC+\angle BCF+\angle CFB=18045+\angle BCF+90=180\angle BCF=45or\angle B{CC}^{\prime }=45\angle B{CC}^{\prime }=\angle B{B}^{\prime }{C}^{\prime }$

Angles made in the same segment of the circle

$\angle A^{{}^{\prime }}{B}^{{}^{\prime }}{C}^{\prime }=\angle B{B}^{\prime }{C}^{\prime }+\angle B{B}^{\prime }{A}^{{}^{\prime }}\angle A^{{}^{\prime }}{B}^{{}^{\prime }}C=45+45=90$

So option $C$ is correct