i. line AB is the tangent to the circle with centre C and radius AC. [Given]

∴ ∠CAB = 90° (i) [Tangent theorem]

ii. seg CA ⊥ line AB [From (i)]

radius = l(AC) = 6 cm

∴ The distance of point C from line AB is 6 cm.

iii. In ∆CAB, ∠CAB = 90° [From (i)]

∴ BC2 = AB2 + AC2 . [Pythagoras theorem]

= 62 + 62

= 2 × 62

∴ BC = \(\sqrt{2 \times 6^{2}}\) [Taking square root of both sides]

= 6 \(\sqrt { 2 }\) cm

∴ d(B, C) = 6 cm

iv. In ∆ABC,

AC = AB = 6cm

∴ ∠ABC = ∠ACB [Isosceles triangle theorem]

Let ∠ABC = ∠ACB =x

In ∆ABC,

∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]

∴ 90° + x + x = 180°

∴ 90 + 2x = 180°

∴ 2x = 180°- 90°

∴ x = \(\frac{90^{\circ}}{2}\)

∴ x = 45°

∴ ∠ABC = 45°