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 GAB=α,  GBC=β,  GCA=γ
let M be the midpoint of BC. Then applying sine rule on ABM, we get
sinB+αsinα=ABBM=2ABBC2sinCsinA=2sinA+BsinAso we have sinB+αsinα=2sinA+BsinAsinB+αsinαsinB=2sinA+BsinBsinAcot α+cot B=2cot B+cot A  cot α=cot B+2cot A

Similarly we get the relations for cot GBC and cot GCA

​Adding up we get, cot α+cot β+cot γ=3cot A +cot B+cot C
Using the fact that if ω is the Brocard Angle, we have
cot ω=cot A+cot B+cot C 
we can write the above relation as
cot α+cot β+cot γ=3cot ω


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