In the given figure, AE is the diameter of the circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.

Join OA, OB, OC, OD.

In △OAB,

OA = OB [Radius of same circle]

∠1 = ∠2.

In △OBC,

OB = OC [Radius of same circle]

∠3 = ∠4.

In △OCD,

OC = OD [Radius of same circle]

∠5 = ∠6.

In △ODE,

OD = OE [Radius of same circle]

∠7 = ∠8.

In △OAB,

⇒ ∠1 + ∠2 + ∠a = 180° [By angle sum property of triangle] ……….(1)

In △OBC,

⇒ ∠3 + ∠4 + ∠b = 180° [By angle sum property of triangle] ……….(2)

In △OCD,

⇒ ∠5 + ∠6 + ∠c = 180° [By angle sum property of triangle] ……….(3)

In △ODE,

⇒ ∠7 + ∠8 + ∠d = 180° [By angle sum property of triangle] ………(4)

Adding (1), (2), (3) and (4) we get,

⇒ ∠1 + ∠2 + ∠a + ∠3 + ∠4 + ∠b + ∠5 + ∠6 + ∠c + ∠7 + ∠8 + ∠d + = 180° + 180° + 180° + 180°

⇒ ∠2 + ∠2 + ∠a + ∠3 + ∠3 + ∠b + ∠6 + ∠6 + ∠c + ∠7 + ∠7 + ∠d + = 720°

⇒ 2∠2 + 2∠3 + 2∠6 + 2∠7 + ∠a + ∠b + ∠c + ∠d = 720°

⇒ 2[∠2 + ∠3] + 2[∠6 + ∠7] + 180° = 720° [As ∠a + ∠b + ∠c + ∠d = 180°]

⇒ 2∠ABC + 2∠CDE = 540°

⇒ ∠ABC + ∠CDE = 270°.

Hence, ∠ABC + ∠CDE = 270°.