Question

Solve the following question with proper theorems :

Answer 1

To prove : ∠AOB = \frac { 1 }{ 2 } (∠C + ∠D)

Proof: In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

\frac { 1 }{ 2 } (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)

Now in ∆AOB

∠1 + ∠2 + ∠3 = 180° ………(iii)

Equating equation (ii) and equation (iii), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + \frac { 1 }{ 2 } (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 + \frac { 1 }{ 2 } (∠C + ∠D)

∠2 = \frac { 1 }{ 2 } (∠C + ∠D)

∠AOB = \frac { 1 }{ 2 } (∠C + ∠D)

Hence proved.

hope it helps.....

Answer 2

Step-by-step explanation:

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