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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=cotA+cotB+cotC

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=cotA+cotB+cotC

Grade:12

2 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
7 years ago
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mycroft holmes
272 Points
7 years ago
Verification isnt the same as proof!!
 
Let \angle GAB = \alpha, \angle GBC = \beta, \angle GCA = \gamma
 
Let M be the midpoint of BC. Then applying sine rule on \triangle ABM, we get
 
\frac{\sin (B+\alpha)}{\sin \alpha} = \frac{AB}{BM} = \frac{2AB}{BC}
 
= \frac{2 \sin C}{\sin A} = \frac{ 2 \sin (A+B)}{\sin A}
 
So we have
 
= \frac{\sin (B+\alpha)}{\sin \alpha} = \frac{ 2 \sin (A+B)}{\sin A}
 
\Rightarrow \frac{\sin (B+\alpha)}{\sin \alpha \sin B} = \frac{ 2 \sin (A+B)}{\sin B\sin A}
 
\Rightarrow \cot \alpha + \cot B = 2 \cot B + \cot A \Rightarrow \cot \alpha = \cot B+ 2 \cot A
 
Similarly we get relations for cot GBC and cot GCA.
 
Adding up we get, \cot \alpha + \cot \beta +\cot \gamma = 3 (\cot A + \cot B+\cot C)
 
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 = 3 \cot \omega

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