Question

In an acute angled $△ABC$, 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

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

Answer 1

Answer: (C)

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

$\angle ADC=90^{\circ }$
$[\because AD$ is altitude $]$
$\angle BAD=45^{\circ }=\angle BA{A}^{\prime }$
Similarly, $\angle BC{C}_1=45^{\circ }$
$\therefore \angle BA{A}_1=\angle B{B}_1{A}_1$
$[\because$ angle on same segment are equal $]$
$\angle BC{C}_1=\angle B{B}_1{C}_1$
$[\because$ angle on same segment are equal $]$
$\therefore \angle A_1{B}_1{C}_1=\angle B{B}_1{A}_1+\angle B{B}_1{C}_1$
$=45^{\circ }+45^{\circ }=90^{\circ }$