Consider the following statements

Statement $\mathrm{I} \cosh ^{-1} x=\tanh ^{-1} x$ has no solution

Statement II $\cosh ^{-1} x=\operatorname{coth}^{-1} x$ has only one solution

The correct answer is

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 $r=$

Two adjacent sides of a triangle are represented by the vectors $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $2 \sqrt{3} \hat{\mathbf{i}}-2 \sqrt{3} \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}$. Then, the least angle of the triangle and perimeter of the triangle are respectively.