KVPY SA Exam 2013 Question Paper

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Question : 8 of 80

Marks: +1, -0

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 ∠ABC = 45° then

∠A_1B_1C_1

equals

45°
60°
90°
135°

Solution:

∠BCH = 45°

= ∠\BCA_1

∠\C_1\CA_1 = ∠\C_1\B_1\A_1 = 90°

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