### Solution of Triangle

goals
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  ?