Solution :

(a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

∠ABC = 180^o – (∠ACB + ∠BAC)….(i)

In the circle, arc AC subtends ∠AOC at

The center and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC …(ii)

Reflex AOC = 2 { (180^o – (ACB + BAC)}

But ∠AOC = 360^o – 2(∠ACB + ∠BAC)

But ∠AOC = 360^o – reflex ∠AOC

=360 – (360^o – 2(∠ACB + ∠BAC)

=360^o – 360^o + 2 (∠ACB + ∠BAC)

=2 (∠ACB + ∠BAC)

Hence ∠AOC = 2 (∠ACB + ∠BAC)

(b) Given : in the figure, O is the center of the circle

To Prove : × + y = z.

Proof : Arc BC subtends ∠AOB at the center and ∠BEC at the remaining part of the circle.

∠BOC = 2 ∠BEC

But ∠BEC = ∠BDC

(Angles in the same segment)

∠BOC = ∠BEC + ∠BDC ……. (ii)

Similarly in ∆ABD

Ext. ∠BDC = x + ∠ABD

= x + ∠EBD ………….(iii)

Substituting the value of (ii) and (iii) in (i)

∠BOC = y – ∠EBD + x + ∠EBD = x + y

Z = x + y