Chứng minh rằng trong tam giác ABC cân nếu: 1 + cosB/ sinB = 2a + c/căn bậc 2(4a^2 - c^2)

* Hướng dẫn giải

$\frac{1+c\text{os}B}{\sin B}=\frac{2\text{a}+c}{\sqrt{4{\text{a}}^2-c^2}}$

$\iff \frac{1+c\text{os}B}{\sin B}=\frac{\sqrt{2\text{a}+c}}{\sqrt{\text{2a}-c}}$

$\iff \frac{{\left(1+c\text{os}B\right)}^2}{{\sin }^{2}B}=\frac{2a+c}{2a-c}$

$\iff \frac{1+c\text{os}B}{1-c\text{os}B}=\frac{2a+c}{2a-c}$

$\iff \left(2a-c\right)\left(1+c\text{os}B\right)=\left(2\text{a}+c\right)\left(1-c\text{os}B\right)$

$\iff 2a+2a\cos B-c-c\cos B=2a-2a\cos B+c-c\cos B$

$\iff 2a\cos B=c\iff 2a\frac{c^2+a^2-b^2}{2ac}=c\iff a=b$ hay $BC=AC$