if G is the centroid of the triangle ABC show that cot GAB+cot GBC+cot GCA= 3cot omega= cot ABG+cotBCG+cotCAG where cot omega=cot A+cot B+cot C

 

Dear Student,

$Let \angle GAB=\alpha , \angle GBC=\beta , \angle GCA=\gamma$
let M be the midpoint of BC. Then applying sine rule on $△ABM$, we get
$$\begin{gathered} \frac{\sin \left(B+\alpha \right)}{\sin \alpha }=\frac{AB}{BM}=\frac{2AB}{BC} \\ \Rightarrow \frac{2\sin C}{\sin A}=\frac{2\sin \left(A+B\right)}{\sin A} \\ so we have \\ \Rightarrow \frac{\sin \left(B+\alpha \right)}{\sin \alpha }=\frac{2\sin \left(A+B\right)}{\sin A} \\ \Rightarrow \frac{\sin \left(B+\alpha \right)}{\sin \alpha \sin B}=\frac{2\sin \left(A+B\right)}{\sin B\sin A} \\ \Rightarrow cot \alpha +cot B=2cot B+cot A \Rightarrow cot \alpha =cot B+2cot A \end{gathered}$$

Similarly we get the relations for cot GBC and cot GCA

​Adding up we get, $cot \alpha +cot \beta +cot \gamma =3\left(cot A +cot B+cot C\right)$
Using the fact that if $\omega$ is the Brocard Angle, we have
$cot \omega =cot A+cot B+cot C$
we can write the above relation as
$cot \alpha +cot \beta +cot \gamma =3cot \omega$

Regards