In ABOC, ∠OCA = 90° (tangent and the radius meet at perpendicular angle) ∠ABO = 90° (same as above) ∵ ∠OCA + ∠ABO = 180°, ABOC is a cyclic quadrilateral
In AOGC, ∠OCA = 90° (proven) ∠OGA is also 90°<span> (line from centre bisects the chord of a circle at a right angle) </span>∵∠OCA = ∠OGA, AOGC is a cyclic quadrilateral (equal angles standing on the same arc on a cyclic quad)
Question (ii): Since ABOC is a cyclic quad, ∠OGF = ∠OAC (exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)
Question (iii): Let ∠OGF = α From (ii), ∠OAC = α
AC = AB <span>(tangents meeting at an external point are equal) AO is common OC = OB (radii)
</span>∠BAO = α (corresponding angles in congruent triangles) ∠BOC = π - 2α <span>(supplementary angles; opposite angles of a cyclic quad are supplementary) </span> It should be pretty simple after this, I'll finish it in a little bit.