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Answer 1

(i)Join BC.

Then, OC = OB = BC triangle. (By construction)

∴ ∠COB is an equilateral triangle.

∴ ∠COB = 60∘.

∴ ∠EOA = 60∘.

(ii)Join CD.

Then, OD = OC = CD (By construction)

∆DOC is an equilateral triangle.

∴ ∠DOC = 60∘.

∴ ∠ FOE = 60∘.

(iii)Join CG and DG.

In ΔODG and ΔOCG,

OD = OC[ Radii of the same arc]

DG=CG [Arcs of equal radii]

OG=OG [Common]

∴ Δ ODG = ΔOCG [SSS Rule]

∴ ∠ DOG= ∠COG [CPCT]

∴ ∠FOG = ∠ EOG = 1/2 ∠FOE = 1/2 (60∘) = 30∘

Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ + 60∘ = 90∘

∴ ∠AOJ= ∠GOJ= 1/2 ∠GOA = ½(90°)=45

Hence justified✅