Question

in a quarilateral ABCD, AO and DO are the bisectors of angle A and ang.D.prove that
ang.AOD=1/2(angB+angC).  pls help me.

Answer 1

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

Answer 2

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

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

$\angle AOD + x + y = 180^o$

$\angle AOD = 180^o - (x+y)$

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

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