# 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 M be the midpoint of BC. Then applying sine rule on $△ABM$, we get

Similarly we get the relations for cot GBC and cot GCA

Using the fact that if $\omega$ is the Brocard Angle, we have