Cho tam giác ABC thỏa mãn $sinA=\frac{sinB+sinC}{cosB+cosC}$ . Chứng minh tam giác ABC là tam giác vuông.

Ta có: $\widehat{A}+\widehat{B}+\widehat{C}=180°$  (tính chất tổng ba góc trong một tam giác)

Suy ra: $\frac{\widehat{A}}{2}+\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}=90°$

⇒ $\frac{\widehat{A}}{2}+\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}=90°$ ⇒ $\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}=90°-\frac{\widehat{A}}{2}$

⇒ $\sin \left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)=\sin \left(90°-\frac{\widehat{A}}{2}\right)=\cos \frac{\widehat{A}}{2}$

$\cos \left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)=\cos \left(90°-\frac{\widehat{A}}{2}\right)=\sin \frac{\widehat{A}}{2}$

$sinA=\frac{sinB+sinC}{cosB+cosC}$⇒ $2sin\frac{\widehat{A}}{2}\cos \frac{\widehat{A}}{2}=\frac{2sin\left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)\cos \left(\frac{\widehat{B}}{2}-\frac{\widehat{C}}{2}\right)}{2\cos \left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)\cos \left(\frac{\widehat{B}}{2}-\frac{\widehat{C}}{2}\right)}$ 

⇒ $2sin\frac{\widehat{A}}{2}\cos \frac{\widehat{A}}{2}=\frac{sin\left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)}{\cos \left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)}$⇔ $2sin\frac{\widehat{A}}{2}\cos \frac{\widehat{A}}{2}=\frac{\cos \left(\frac{\widehat{A}}{2}\right)}{\sin \left(\frac{\widehat{A}}{2}\right)}$

⇔ $2si{n}^2\left(\frac{\widehat{A}}{2}\right)=1$

⇔ $1-2si{n}^2\left(\frac{\widehat{A}}{2}\right)=0$ 

⇔ $\cos \left(\widehat{A}\right)=0$

⇔ $\widehat{A}=90°$

Vậy tam giác ABC vuông.