Question

Two fixed circles in a plane intersect in points P and Q. A variable line through P meets the circles again in A and B. Prove that the angle AQB is of constant measure.

Answer 1

Let the line A′B′ be another line through P meeting the circles in A′ and B′.

Given, APB is a line through P meeting the circle in A and B respectively.

∠PAQ = ∠PA′Q {Angles in the same segment are equal} 

∠PBQ = ∠PB′Q 

Now in ∆ AQB, 

∠AQB = 180° – (∠QAB + ∠QBA) 

= 180° – (∠PAQ + ∠QBP) 

= 180° – (∠PA’Q + ∠PB’Q) 

= ∠A′Q B′

∴ ∠AQB is the same for all lines APB. Thus, ∠AQB is a constant angle.