another problem from a book
$ABCD$ convex quadrilateral,$C(O,r)$ circle such that $O\in [AB]$ and touched $BC,DC,DA$ in $E,F,N$ respectively
Prove that when a quadrilateral is to be drawn within a circle, then
$$AB=DA+BC$$
solution: let $E,F,N$ touched circle in $BC,DC,DA$ respectively
then : $EB=r\cot B,AN=r\cot A,EC=r\cot (\frac{C}{2})=r\tan (\frac{A}{2}),ND=r\cot (\frac{D}{2})=r\tan (\frac{B}{2})$ then $AD+BC=r(\cot A+\cot B+\tan (\frac{A}{2})+\tan (\frac{B}{2})=r(\frac{1}{\sin A}+\frac{1}{\sin B})=AB$
note that : $OB=\frac{r}{\sin B},OA=\frac{1}{\sin A}$ I don't understand this solution actually for me new relations ? can someone explain to me ?
The crucial part of this proof concerns the fact that $$\cot A+\cot B+\tan (\frac{A}{2})+\tan (\frac{B}{2})=\frac{1}{\sin A}+\frac{1}{\sin B}.$$
This follows since $$\cot A+\tan (\frac{A}{2})=\frac{1-2(\sin A/2)^2}{\sin A}+ \frac{(2\sin A/2)^2}{\sin A} =\frac{1}{\sin A}$$ with the identical result for $B$.