Q)  In the given figure AC is the diameter of the circle with centre O. CD is parallel to BE. angle AOB = 80 deg and angle ACE = 20 deg .

Calculate:
(a) angle BEC
(b) angle BCD
(c) angle CED

Ans: 

(a) ∠ BEC:

Step 1: Given that ∠ AOB = 80⁰

∠ BOC = 180 – ∠ AOB

∴ ∠ BOC = 180 – 80 = 100⁰

Step 2: Since, Angle subtended by a cord at the center by a cord is two times of the angle subtended by that cord at the circumference.

∴ ∠ BOC = 2 x ∠ BEC

∴ ∠ BEC = ∠ BOC

∴ ∠ BEC = x 100⁰

∴ ∠ BEC = 50⁰

Therefore, ∠ BEC is 50⁰

(b) ∠ BCD:

Step 3: Let’s connect AB 

∴ ∠ AOB = 2 x ∠ ACB    (as explained in step 2)

∴ ∠ ACB = x ∠ AOB

∴ ∠ ACB = x 80⁰

∴ ∠ ACB = 40⁰

Step 4: Since CD an BE are parallel (given) and CE is the transversal cutting these lines,

∴ ∠ BEC = ∠ ECD   (alternate interior angles)

∴ ∠ ECD = 50⁰

Given that ∠ ACE = 20⁰

Step 5: ∠ BCD = ∠ ACB + ∠ ACE + ∠ ECD

∴ ∠ BCD = 40⁰ + 20⁰ + 50⁰

∴ ∠ BCD = 110⁰

Therefore, ∠ BCD is 110⁰

(c) ∠ CED: 

Step 6: In Δ BCE, ∠ ACB  = 40⁰   (from step 3) 

and ∠ ACE = 20⁰

∠ BCE = ∠ ACE + ∠ ACB

∠ BCE = 20⁰ + 40⁰ = 60⁰

∠ BEC = 50⁰                 (from step 2)

Step 7: ∴ ∠ BCE + ∠ BEC + ∠ EBC = 180⁰

∴ ∠ BCE + ∠ BEC + ∠ EBC = 180⁰

∴ 60⁰ + 50⁰ + ∠ EBC = 180⁰

∴ ∠ EBC = 180⁰ – 110⁰

∴ ∠ EBC = 70⁰

Step 8: ∠ EDC + ∠ EBC = 180⁰

(since in a cyclical quadrilateral, opposite angles are complimentary)

∴ ∠ EDC + 70⁰ = 180⁰

∴ ∠ EDC = 180⁰ – 70⁰

∴ ∠ EDC = 110⁰

Step 9: In Δ DCE,  ∠ EDC + ∠ DCE + ∠ DEC = 180⁰

∠ EDC = 110⁰                      (from step 8)

∠ ECD = ∠ DCE = 50⁰         (from step 4)

∴ 110⁰ + 50⁰ + ∠ DEC = 180⁰

∴ ∠ DEC = 180⁰ – 160⁰

∴ ∠ DEC = 20⁰

Therefore, ∠ DEC is 20⁰

Please press the “Heart” button if you like the solution.