[tex3]cos(A)+cos(B) = 2cos\left(\frac{A+B}{2}\right)\cdot cos\left(\frac{A-B}{2}\right) =2cos\left(\frac{\pi-C}{2}\right)\cdot cos\left(\frac{A-B}{2}\right) =[/tex3]
[tex3]2sen\left(\frac{C}{2}\right)\cdot cos\left(\frac{A-B}{2}\right) [/tex3]
- [tex3]sen\left(\frac{C}{2}\right)\neq 0[/tex3]
, do contrário C=0° ou C=360°
[tex3]2sen\left(\frac{C}{2}\right)\cdot cos\left(\frac{A-B}{2}\right) = 4sen^2\left(\frac{C}{2}\right) [/tex3]
[tex3]\frac{cos\left(\frac{A-B}{2}\right)}{sen\left(\frac{C}{2}\right) } = 2 [/tex3]
- ma.png (30.57 KiB) Exibido 5492 vezes
Teorema da bissetriz interna:
[tex3]\frac{b}{sen(c-t)} = \frac{a}{t}\rightarrow t = \frac{ac}{b+a}[/tex3]
Lei dos senos:
[tex3]\frac{t}{sen(\frac{C}{2})} = \frac{a}{sen(a+\frac{c}{2})}[/tex3]
[tex3]\frac{b+a}{c} = \frac{sen(a+\frac{c}{2})}{sen\left(\frac{c}{2}\right)}[/tex3]
Mas: [tex3]a+\frac{c}{2} - (\frac{a-b}{2}) = \frac{a+b+c}{2} = 90^o[/tex3]
Portanto:
[tex3]\frac{b+a}{c} = \frac{sen(a+\frac{c}{2})}{sen\left(\frac{c}{2}\right)}=\frac{cos\left(\frac{A-B}{2}\right)}{sen\left(\frac{C}{2}\right) } = 2[/tex3]