If the angular bisector of the angle $A$ of the $\triangle A B C$ meets its circumcircle at $E$ and the opposite side $B C$ at $D$, then $D E \cos \frac{A}{2}=$

In a $\triangle A B C, a=5, b=4$ and $\tan \frac{C}{2}=\sqrt{\frac{7}{9}}$, then its inradius $r=$

$y-x=0$ is the equation of a side of a $\triangle A B C$. The orthocentre and circumcentre of the $\triangle A B C$ are respectively $(5,8)$ and $(2,3)$. The reflection of orthocentre with respect to any side of the triangle lies on its circumcircle. Then, the radius of the circumcircle of the triangle is

If $a=3, b=5, c=7$ are the sides of a $\triangle A B C$, then $\cot A+\cot B+\cot C=$