AO and DO a...

Question

$A O$ and $D O$ are the bisectors of $∠ A$ and $∠ D$ of the quadrilateral $A B C D$ . Prove that $∠ A O D = \frac{1}{2} (∠ B + ∠ C)$

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Solution

To proof:
$\begin{matrix} ∠ A O D = \frac{1}{2} (∠ B + ∠ C) p r o o f : ∠ A + ∠ B + ∠ C + ∠ D = 360^{0} [a n g l e o f a q u a r d .] = \frac{1}{2} ∠ A + \frac{1}{2} ∠ B + \frac{1}{2} ∠ C + \frac{1}{2} ∠ D = 180^{0} = ∠ O A D + ∠ D A D = 180^{0} - \frac{1}{2} (∠ B + ∠ C) . . . . . . . . . . . . . (1) ∵ ∠ O D A + ∠ D A D + ∠ A O D = 180^{0} - ∠ A O D P u t t i n g t h i s v a l u e i n (1) 180^{0} - ∠ A O D = 180^{0} - \frac{1}{2} (∠ B + ∠ C) ∠ A O D = \frac{1}{2} (∠ B + ∠ C) \end{matrix}$

Hence, proved.

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