Solution of Triangle

In triangle ABC, prove that the area of the in-circle is to the area of the triangle itself $$=\pi : \cot (A/2) \cot (B/2) \cot (C/2).$$
If the coordinates of the vertices of the triangle $$ABC$$ be $$(-1,6),(-3,-9)$$, and $$(5,-8)$$ respectively, then the equation of the median through $$C$$ is
If in a $$\triangle ABC, \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$$, then $$\triangle ABC$$ is
If O is the circumcentre and O' is the orthocentre of a triangle ABC and if $$\overline { A P }$$ is the circumdiameter then $$\overline { A O ^ { \prime } } + \overline { O ^ { \prime } B } + \overline { O ^ { \prime } C } =$$ ?
The area of cyclic quadrilateral $$ABCD$$ is $$\dfrac{3\sqrt{3}}{4}$$. The radius of the circle circumscribing it is $$1$$.If $$ AB =1$$ $$BC=\sqrt{3}$$ then $$\angle ABC$$ is equal to  ?