Question

Na figura, AQ e AP são, respectivamente bissetrizes interna e externa do triângulo ABC. Se BQ = 8m e QC = 6m, então, a medida de QP, em metros é?

Answer 1

Olá Marco!

Sejam $\mathbf{\overline{AB} = a}$ e $\mathbf{\overline{AC} = b}$.

Então:

- BISSETRIZ INTERNA (AQ):

$\displaystyle \\ \mathsf{\frac{\overline{AB}}{\overline{BQ}} = \frac{\overline{AC}}{\overline{QC}}} \\\\\\ \mathsf{\frac{a}{8} = \frac{b}{6}} \\\\\\ \mathsf{\frac{a}{b} = \frac{8}{6}} \\\\\\ \mathsf{\frac{a}{b} = \frac{4}{3}}$

 Considere "k" um inteiro não negativo e não nulo, ou seja, $\mathbf{k \in \mathbb{Z}_{+}^{\ast}}$. Daí,

$\\ \displaystyle \mathsf{\frac{a}{b} = \frac{4k}{3k} \Rightarrow \ \begin{cases} \mathsf{a = 4k} \\ \mathsf{b = 3k} \end{cases}}$


 
- BISSETRIZ EXTERNA (AP):

$\displaystyle \\ \mathsf{\frac{\overline{AB}}{\overline{BP}} = \frac{\overline{AC}}{\overline{CP}}} \\\\\\ \mathsf{\frac{a}{8 + 6 + \overline{CP}} = \frac{b}{\overline{CP}}} \\\\\\ \mathsf{\frac{a}{14 + \overline{CP}} = \frac{b}{\overline{CP}}}$


 Substituindo "a" e "b",

$\displaystyle \\ \mathsf{\frac{a}{14 + \overline{CP}} = \frac{b}{\overline{CP}}} \\\\\\ \mathsf{\frac{4k}{14 + \overline{CP}} = \frac{3k}{\overline{CP}}} \\\\\\ \mathsf{\frac{4}{14 + \overline{CP}} = \frac{3}{\overline{CP}}} \\\\\\ \mathsf{4 \cdot \overline{CP} = 3 \cdot 14 + 3 \cdot \overline{CP}} \\\\ \boxed{\mathsf{\overline{CP} = 42 \ m}}$


 Por fim, temos que:

$\displaystyle \\ \mathsf{\overline{QP} = \overline{QC} + \overline{CP}} \\\\ \mathsf{\overline{QP} = 6 + 42} \\\\ \boxed{\boxed{\mathsf{\overline{QP} = 48 \ m}}}$