Question

Prove that:

i. ∠BAT = ∠BPA

ii. ∠BAS = ∠AQB

Answer 1

(Sum of two acute angles of a right ∆ABC)

∠CAT = 90°

(Diameter is to the tangent at the point of contact) ∠CAB + ∠BAT = 90°

∠CAB + ∠BCA = ∠CAB + ∠BAT

(from (1) and (2)) ∠BCA = ∠BAT

∠BCA = ∠BPA

(angles in the same segment)

∠BAT = ∠BPA (from (3) and (4) (5))

ii. Now ∠BPA + ∠AQB = 180°

(opposite angles of a cyclic quadrilateral) (from (5) and (6))

Also ∠BAT + ∠BAS = ∠BAT + ∠BAS

(from (6) and (7)) ∠BAS = ∠AQB