In the adjoining diagram ∠DOE = 46° and ∠BGH = 113°.

We know that,

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠DBE = $\frac{1}{2}$ DOE

⇒ ∠DBE = $\frac{1}{2}$ × 46° = 23°

From figure,

⇒ ∠DBC = ∠DBE = 23°.

From figure,

⇒ ∠BDA = 90° [Angle in a semicircle]

⇒ ∠BDC = 180° - ∠BDA [Angles on a straight line]

⇒ ∠BDC = 180° - 90°

⇒ ∠BDC = 90°.

In triangle BDC,

Sum of angles in a triangle = 180°

⇒ ∠BDC + ∠DCB + ∠DBC = 180°

⇒ 90° + ∠DCB + 23° = 180°

⇒ 113° + ∠DCB = 180°

⇒ ∠DCB = 180° - 113°

⇒ ∠DCB = 67°

From figure,

∠DCE = ∠DCB = 67°.

Hence, ∠DBC = 23° and ∠DCE = 67°.

(b) From figure,

⇒ ∠BCH = ∠DCE = 67° [Common angles]

⇒ ∠BGH = 113°

⇒ ∠BGH + ∠BCH = 113° + 67° = 180°.

Since, the sum of opposite sides of quadrilateral is 180°, thus CBGH is a cyclic quadrilateral.

Hence, proved that CBGH is a cyclic quadrilateral.