Question

In an acute angled $\triangle A B C$, the altitudes from $A, B, C$ 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

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

Answer 1

Answer: (C)

Given, $A B C$ is an acute angle triangle.
$\angle B=45^{\circ}$

$\angle A D C=90^{\circ}$
$[\because A D$ is altitude $]$
$\angle B A D=45^{\circ}=\angle B A A^{\prime}$
Similarly, $\angle B C C_{1}=45^{\circ}$
$\therefore \angle B A A_{1}=\angle B B_{1} A_{1}$
$[\because$ angle on same segment are equal $]$
$\angle B C C_{1}=\angle B B_{1} C_{1}$
$[\because$ angle on same segment are equal $]$
$\therefore \angle A_{1} B_{1} C_{1} =\angle B B_{1} A_{1}+\angle B B_{1} C_{1} $
$=45^{\circ}+45^{\circ}=90^{\circ}$