Arrange the sides of △ BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.

From figure,

DAC is a straight line.

∴ ∠DAB + ∠BAC = 180°

⇒ 137° + ∠BAC = 180°

⇒ ∠BAC = 180° - 137° = 43°.

EBC is a straight line.

∴ ∠EBA + ∠ABC = 180°

⇒ 106° + ∠ABC = 180°

⇒ ∠ABC = 180° - 106° = 74°.

In △ ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

⇒ 74° + 43° + ∠ACB = 180°

⇒ 117° + ∠ACB = 180°

⇒ ∠ACB = 180° - 117° = 63°.

Since, OB is the bisector of angle ABC.

∴ ∠OBC = $\dfrac{∠ABC}{2} = \dfrac{74°}{2}$ = 37°.

Since, OC is the bisector of angle ACB.

∴ ∠OCB = $\dfrac{∠ACB}{2} = \dfrac{63°}{2}$ = 31.5°.

In △ OBC,

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 37° + 31.5° + ∠BOC = 180°

⇒ 68.5° + ∠BOC = 180°

⇒ ∠BOC = 180° - 68.5° = 111.5°

∴ ∠BOC > ∠OBC > ∠OCB

∴ BC > CO > BO (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Hence, sides of triangle BOC in descending order are BC > CO > BO.