Question

PLEASE ANSWER FAST AND SHOW FULL SOLUTION!!

Answer 1

given: ∠ACE = 50°
construction: join; BE,  AC,  OC,  OD
find: (a)∠CBE (b)∠CDE (c)∠AOB
prove that BOIICE
SOLUTION: 
2∠AEC = ∠AOC ⇒ ∠AOC = 100°
In ΔAOB and ΔBOC, AB = BC, AO = OC and OB = OB
∴ ∠BOA = ∠BOC
⇒∠BOA = 50° and ∠BOC = 50°
⇒ ∠BOA = ∠CEO 
∴ BOIICE [If corresponding angles are equal, the lines are parallel]
now, ∠AOC + ∠COE = 180°
∴ ∠COE = 80°

   ∠COE = 2∠CBE
∴ ∠CBE = 40°
In cyclic quadrilateral BCDE,
           ∠CBE + ∠CDE = 180°
                         ∠CDE = 140°