In a quadrilateral, CO and DO are the bisectors of ∠C and ∠D respectively.

Prove that $∠COD=\frac{1}{2}(∠A + ∠B)$."

Given: In a quadrilateral, CO and DO are the bisectors of ∠C and ∠D respectively.

To prove: Here we have to prove that $\angle \mathrm{COD} \ =\ \frac{1}{2} (\angle \mathrm{A} \ +\ \angle B)$.

Solution:

∠A $+$ ∠B $+$ ∠C $+$ ∠D = 360^o 

∠C $+$ ∠D = 360^o $-$ (∠A $+$ ∠B) 

In △COD,

∠COD $+$ ∠1 $+$ ∠2 = 180^o  

∠COD = 180^o $−$ (∠1 $+$ ∠2)

∠COD = 180^o $−$ $\frac{1}{2}$(∠C $+$ ∠D)

∠COD = 180^o $−$ $\frac{1}{2}$[360^o $−$ (∠A $+$ ∠B)]

∠COD = $\frac{1}{2}$(∠A $+$ ∠B)

So, it is proved that ∠COD = $\frac{1}{2}$(∠A $+$ ∠B).

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Updated on: 10-Oct-2022

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