- A$2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
- B$4\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
- ✓$\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
- D$8\,\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
$\therefore \,\frac{A}{2} + \frac{B}{2} = {90^o} - \frac{C}{2}$
$\therefore \cot \left( {\frac{A}{2} + \frac{B}{2}} \right) = \cot \left( {{{90}^o} - \frac{C}{2}} \right)$
or $\frac{{\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\, - 1}}{{\cot \frac{B}{2}\, + \,\cot \frac{A}{2}}} = \tan \frac{C}{2} = \frac{1}{{\cot \frac{C}{2}}}$
or $\left( {\cot \frac{A}{2}\cot \frac{B}{2} - 1} \right)\cot \frac{C}{2} = \cot \frac{B}{2} + \cot \frac{A}{2}$
$\cot \frac{A}{2}\,.\,\cot \frac{B}{2}\,.\,\cot \frac{C}{2} = \cot \frac{C}{2} + \cot \frac{B}{2}$ $ + \cot \frac{A}{2}.$
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$(A)$ The length of the line segment $O A_1$ is $4 \sqrt{3}$
$(B)$ The length of the line segment $A_1 B_1$ is 16
$(C)$ The orthocenter of the triangle $A_1 B_1 C_1$ is $(0,0)$
$(D)$ The orthocenter of the triangle $A_1 B_1 C_1$ is $(1,0)$