Demonstrações ⇒ (Demonstração) Teorema da Bissetriz Interna

[CarlosBruno]

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[tex3]\text{Considere: }\overline{BL}=m \quad \overline{LC}=n \quad \overline{AB}=c \quad \overline{AC}=b \quad \angle ALB=\beta \quad \text{ e } \quad \angle BAL=\angle LAC=\alpha\\ \text{Aplicando lei dos senos no } \triangle \text{BAL } \implies \frac{m}{\sen\ \alpha}=\frac{c}{\sen\ \beta}\therefore \frac{\sen\ \alpha}{\sen \ \beta}=\frac{m}{c} \text{ (I)} \\ \text{Aplicando lei dos senos no } \ \triangle \text{CAL}\implies \frac{n}{\sen\ \alpha}=\frac{b}{\sen\ \pi-\beta}\therefore \frac{\sen\ \alpha}{\sen \ \beta}=\frac{n}{b}\text{ (II)}\\ \therefore \text{(I) = (II)}\implies \frac{m}{c}=\frac{n}{b}\Rightarrow mb=nc\Rightarrow \frac{b}{n}=\frac{c}{m} \qquad CQD[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\text{Considere: }\overline{BL}=m \quad \overline{LC}=n \quad \overline{AB}=c \quad \overline{AC}=b \quad \angle ALB=\beta \quad \text{ e } \quad \angle BAL=\angle LAC=\alpha\\ \text{Aplicando lei dos senos no } \triangle \text{BAL } \implies \frac{m}{\sen\ \alpha}=\frac{c}{\sen\ \beta}\therefore \frac{\sen\ \alpha}{\sen \ \beta}=\frac{m}{c} \text{ (I)} \\ \text{Aplicando lei dos senos no } \ \triangle \text{CAL}\implies \frac{n}{\sen\ \alpha}=\frac{b}{\sen\ \pi-\beta}\therefore \frac{\sen\ \alpha}{\sen \ \beta}=\frac{n}{b}\text{ (II)}\\ \therefore \text{(I) = (II)}\implies \frac{m}{c}=\frac{n}{b}\Rightarrow mb=nc\Rightarrow \frac{b}{n}=\frac{c}{m} \qquad CQD[/tex3]