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Grade: 12

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


4 years ago

## Answers : (2)

Sourabh Singh
IIT Patna
askIITians Faculty
2104 Points


4 years ago
mycroft holmes
272 Points
							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 $\cot \alpha + \cot \beta +\cot \gamma = 3 \cot \omega$

4 years ago
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