Using the informations given in the adjoining figure, write the sides of △BOC in ascending order of length.

Since BAD is a straight line,

⇒ ∠BAC + ∠CAD = 180°

⇒ ∠BAC + 135° = 180°

⇒ ∠BAC = 180° - 135° = 45°

From the figure,

∠BCO = ∠OCA = x (assume)

Since BCE is a straight line,

⇒ ∠BCO + ∠OCA + ∠ACE = 180°

⇒ x + x + 95° = 180°

⇒ 2x + 95° = 180°

⇒ 2x = 180° - 95° = 85°

⇒ x = $\dfrac{85°}{2}$ = 42.5°

∠BCA = 2x = 85°

From the figure,

∠ABO = ∠OBC = y (assume)

∠CBA = 2y

In Δ ABC, sum of all angles of the triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 45° + 2y + 85° = 180°

⇒ 130° + 2y = 180°

⇒ 2y = 180° - 130°

⇒ 2y = 50°

⇒ y = $\dfrac{50°}{2}$

⇒ y = 25°

In Δ BOC, sum of all angles of the triangle is 180°.

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 25° + 42.5° + ∠BOC = 180°

⇒ 67.5° + ∠BOC = 180°

⇒ ∠BOC = 180° - 67.5°

⇒ ∠BOC = 112.5°

Now, in triangle OBC,

∠BOC > ∠OCB > ∠OBC

Side opposite to greater angle is greater.

Hence, OC < OB < BC.