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Bình phương 2 vế ta đc:
$\frac{(1+cosB)^2}{sin^2B}=\frac{2a+c}{2a-c}\Leftrightarrow \frac{(1+cosB)^2}{(1-cosB)(1+cosB)}=\frac{2a+c}{2a-c}$
$\Leftrightarrow \frac{1+cosB}{1-cosB}=\frac{2a+c}{2a-c}\Leftrightarrow \frac{a^2+c^2-b^2+2ac}{2ac-a^2-c^2+b^2}=\frac{2a+c}{2a-c}$
$\Leftrightarrow 4a^3=4ab^2\Leftrightarrow a=b\Leftrightarrow BC=AC\Rightarrow \Delta $ABC cân tại C
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