In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.

From figure,

∠BEC = ∠BDC (∵ angles in same segment are equal)

Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∴ ∠BOC = 2∠BEC (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

⇒ ∠BOC = ∠BEC + ∠BEC = ∠BEC + ∠BDC

⇒ ∠BOC = ∠BEC + ∠BDC …..(i)

In △OBE,

⇒ ∠EOD = ∠EBO + ∠BEO ….(ii) (∵ exterior angle is equal to the sum of two opposite interior angles.)

From figure,

∠EBO = ∠EBD and ∠BEO = ∠BEC and Exterior angle ∠EOD = y.

Putting these values in eqn (ii) we get,

⇒ y = ∠EBD + ∠BEC
⇒ ∠BEC = y - ∠EBD …..(iii)

In △ABD,

∠BDC = ∠BAD + ∠ABD ….(iv) (∵ exterior angle is equal to the sum of two opposite interior angles.)

From figure,

∠ABD = ∠EBD and ∠BAD = x.

Putting these values in eqn (iv) we get,

∠BDC = x + ∠EBD ……(v)

Putting value of ∠BEC and ∠BDC from eqn (iii) and (v) respectively in (i) we get,

⇒ ∠BOC = ∠BEC + ∠BDC
⇒ ∠BOC = y - ∠EBD + x + ∠EBD
⇒ ∠BOC = x + y

From figure,

∠BOC = z.

∴ z = x + y.

Hence, proved that x + y = z.