Answers

  • Brainly User
2014-09-12T14:10:17+05:30
In triangle AOD,  angle AOD = 180 -(1/2)(angle A + Angle D) -------(1)

sum of angles of a quadrilateral is 360, hence
Angle A + Angle D + Angle B + Angle C = 360
==> (1/2)(Angle A + Angle D + Angle B + Angle C) = 360/2 = 180
==> (1/2)(Angle A + Angle D) + (1/2)(Angle B + Angle C) = 180
==>  (1/2)(Angle B + Angle C) = 180 - (1/2)(Angle A + Angle D)
                                            = Angle AOD from equation (1) above
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2014-09-12T15:19:19+05:30
In quad. ABCD
\angle A + \angle B + \angle C + \angle D = 360^o

2x + \angle B + \angle C + 2y = 360^o

2x + 2y = 360^o -  (\angle B + \angle C)

x + y = 180^o -   \frac{(\angle B + \angle C)}{2}   ........(1)

Now, consider ΔAOD

\angle AOD + \angle ODA + \angle OAD = 180^o

\angle AOD + x + y = 180^o

\angle AOD = 180^o - (x+y) 

\angle AOD = 180^o - 180^o + \frac{(\angle B + \angle C)}{2}

\angle AOD = \frac{(\angle B + \angle C)}{2}


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