In the given figure, $ O C $ and $ O D $ are the angle bisectors of $ \angle B C D $ and $ \angle A D C $ respectively. If $ \angle A=105^{\circ} $, find $ \angle B $.
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Given: In the given figure, $OC$ and $OD$ are the angle bisectors of $\angle BCD$ and $\angle ADC$ respectively and $\angle A=105^{\circ}$. 

To do: To find $\angle B$.

Solution:

$\because OD$ is the bisector of $ADC$

$\therefore OC$ is the bisector of $BCD$

$\Rightarrow \angle ADC=\angle D=\angle ODC\times 2$

$\Rightarrow \angle D=25^o\times 2=50^o$

Similarly: $\angle BCD=\angle C=OCD\times 2$

$\Rightarrow \angle C=30^o\times 2=60^o$

It is known that $A+B+C+D=360^o$    $( sum\ of\ angles\ of\ quadrilateral=360^o )$

$\Rightarrow 105^o+B+60^o+50^o=360^o$

$\Rightarrow 105^o+B+110^o=360^o$

$\Rightarrow B+105^o+110^o=360^o$

$\Rightarrow B+215^o=360^o$

$\Rightarrow B=360^o - 215^o$

$\Rightarrow B=145^o$

Thus, $\angle B=145^o$.