Q3) In a quadrilateral DARE , OA and OR are respectively the bisectors of ∠A and ∠R. Prove that ∠AOR = ½(∠D and ∠E).
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In quadrilateral [math]ABCD, CO[/math] and [math]DO[/math] are bisectors of [math]\angle C[/math] and [math]\angle D[/math] respectively. How would one prove that [math]\angle COD=\frac{1}{2}\angle A+\frac{1}{2}\angle B?[/math]
In [math]\triangle COD, \angle COD+\angle ODC+\angle OCD=180^o.[/math]
[math]\Rightarrow\qquad \angle COD=180^o-\angle ODC-\angle OCD.[/math]
[math]CO[/math] and [math]DO[/math] are bisectors of [math]\angle C[/math] and [math]\angle D.[/math]
[math]\Rightarrow\qquad \angle ODC=\frac{1}{2}\angle D[/math] and [math]\angle OCD=\frac{1}{2}\angle C.[/math]
[math]\Rightarrow\qquad \angle COD=180^o-\frac{1}{2}\angle D-\frac{1}{2}\angle C.[/math]
In quadrilateral [math]ABCD, \angle A+\angle B+\angle C+\angle D=360^o.[/math]
[math]\Rightarrow\qquad \frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C+\frac{1}{2}\angle D=180^o.[/math]
[math]\Rightarrow\qquad \frac{1}{2}\angle A+\frac{1}{2}\angle B=180^o-\frac{1}{2}\angle C-\frac{1}{2}\angle D.[/math]
But [math]\angle COD=180^o-\frac{1}{2}\angle D-\frac{1}{2}\angle C.[/math]
[math]\Rightarrow\qquad \angle COD=\frac{1}{2}\angle A+\frac{1}{2}\angle B.[/math]