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

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Please find below the solution to the asked query:

Draw the figureLetGAB=x and GBC=y and GCA=zLet M be the mid point of BC. Applying sine rule in ABMsinB+xsinx=ABBM=2ABBC=2sinCsinA=2sinπ-A+BsinA=2sinA+BsinAi.e.sinB+xsinx=2sinA+BsinAsinB+xsinx.sinB=2sinA+BsinA.sinBsinB.cosx+cosB.sinxsinx.sinB=2sinA.cosB+2cosA.sinBsinA.sinBsinB.cosxsinx.sinB+cosB.sinxsinx.sinB=2sinA.cosBsinA.sinB+2cosA.sinBsinA.sinBcotx+cotB=2cotB+2cotAcotx=cotB+2cotAOn similar note we get cot y and cot zOn adding 3 equations we havecot x+cot y+cot z=3cotA+cotB+cotCGiven that cotA+cotB+cotC=cotωcot x+cot y+cot z=3cotωcotGAB+cotGBC+cotGCA=3cotω

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