right triangle if a+c=b ctg (<B/2) (2016 Moldova NMO 11.6)

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right triangle if a+c=b ctg (<B/2) (2016 Moldova NMO 11.6)

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#1 h Mar 19, 2021, 4:15 PM

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Let $a, b, c$ be the lengths of the sides of the triangle $ABC$ , and $\alpha, \beta, \gamma$ the measures of opposite angles respectively. Show that if $a+c=b ctg\frac{\beta}{2}$ , then the triangle is right.

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#2 h Mar 19, 2021, 7:45 PM • 1 Y

Y by Mango247

$a+c=b \cdot \cot \frac{\beta}{2}$ .

$\sin \alpha+\sin \gamma=\sin \beta \cdot \frac{\cos \frac{\beta}{2}}{\sin \frac{\beta}{2}}$ .
$2\sin \frac{\alpha+\gamma}{2}\cos \frac{\alpha-\gamma}{2}=2\sin \frac{\beta}{2}\cos \frac{\beta}{2} \cdot \frac{\cos \frac{\beta}{2}}{\sin \frac{\beta}{2}}$ .
$2\cos \frac{\beta}{2}\cos \frac{\alpha-\gamma}{2}=2\cos^{2}\frac{\beta}{2}$ .
$\cos \frac{\alpha-\gamma}{2}=\cos \frac{\beta}{2}$ .

1) $\frac{\alpha-\gamma}{2}=\frac{\beta}{2}$ or $\alpha=\beta+\gamma$ or $\alpha=180^{\circ}-\alpha$ or $\alpha=90^{\circ}$ .
2) $\frac{\alpha-\gamma}{2}=-\frac{\beta}{2}$ or $\gamma=90^{\circ}$ .

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