Question

In the figure, ∠BAD = 40° then find ∠BCD.

Answer 1

‘O’ is the centre of the circle. 

∴ In ΔOAB; OA = OB (radii) 

∴ ∠OAB = ∠OBA = 40° 

(∵ angles opp. to equal sides) 

Now ∠AOB = 180° – (40° + 40°)

 (∵ angle sum property of ΔOAB) 

= 180°-80° = 100°

But ∠AOB = ∠COD = 100° 

Also ∠OCD = ∠ODC [OC = OD] 

= 40° as in ΔOAB 

∴ ∠BCD = 40° 

(OR) 

In ΔOAB and ΔOCD 

OA = OD (radii) 

OB = OC (radii) 

∠AOB = ∠COD (vertically opp. angles) 

∴ ΔOAB ≅ ΔOCD 

∴ ∠BCD = ∠OBA = 40° 

[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]