2b^2=a^2+c^2 => 2cotB=cotA+cotC

**james_bond** · October 9th 2009, 11:09 PM

Prove that if $2b^2=a^2+c^2$ in a triangle then $2\cot\beta=\cot\alpha+\cot\gamma$ .

**red_dog** · October 10th 2009, 01:18 AM

I'll use the letters A, B, C for the angles of the triangles instead of $\alpha, \beta, \gamma$ .

We have $b^2=a^2+c^2-2ac\cos B$ . Then

$2a^2+2c^2-4ac\cos B=a^2+c^2$

$a^2+c^2=4ac\cos B$

$4R^2\sin^2A+4R^2\sin^2C=16R^2\sin A\sin C\cos B$

Divide by $4R^2$ :

$\sin^2A+\sin^2C=4\sin A\sin C\cos B$

$\sin^2A-\sin A\sin C\cos B+\sin^2C-\sin A\sin C\cos B=2\sin A\sin C\cos B$

Using that $\sin A=\sin(B+C), \ \sin C=\sin(A+B)$ we have

$\sin A(\sin(B+C)-\sin C\cos B)+\sin C(\sin(A+B)-\sin A\cos B)=2\sin A\sin C\cos B$

Now, using the formula $\sin(B+C)=\sin B\cos C+\sin C\cos B$ and $\sin(A+B)=\sin A\cos B+\sin B\cos A$ we have

$\sin A\sin B\cos C+\sin C\sin B\cos A=2\sin A\sin C\cos B$

Divide all terms by $\sin A\sin B\sin C$ and we get

$\cot C+\cot A=2\cot B$

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