Question

In the fig. angle ADC = 130° and chord BC= chord BE. Find angle CBE.

Answer 1

The point A, B, C and D formed a cyclic quadrilateral .

∴ ∠ADC + ∠OBC = 180°  

⇒ 130°+ ∠OBC = 180°    

⇒ ∠OBC = 180°   

Now, in ΔBOC and ΔBOE , 

BC = BE [given]

OC = OE  [radii of the same circle]

OB = OB   [common side]

∴ ΔBOC ≅ ΔBOE [by SSS congruent rule]

Then , ∠OBC = ∠OBE = 50° [CPCT]

∴ ∠CBE = ∠CBO + ∠EBO =  50° +  50° = 100°

Answer 2

Answer:

∠CBE =100°

Step-by-step explanation:

In △BCO and △BEO

⇒  BC=BE                 [ Given ]

⇒  ∠BCO=∠BEO              [ Base angles of equal sides are also equal ]

⇒  BO=BO                  [ Common side ]

∴  △BCO≅△BEO             [ By SAS congruence rule ]

⇒  ∠CBO=∠OBE        [ C.P.C.T. ]

Quadrilateral ABCD is a cyclic quadrilateral since, its points lies on a circle.

⇒  ∠ADC+∠CBA=180°

[ Sum of opposite angles of a cyclic quadrilateral is supplementary ]

⇒  130° +∠CBA=180°

⇒  ∠CBA=50°

 i.e. ∠CBO=50°

, ∠OBE=50°

⇒  ∠CBE=50° +50° =100°

Therefore, ∠CBE =100°