Let the line $L_{1}:x + 3 = 0$ intersect the lines $L_{2}:x - y = 0$ and $L_{3}:3x + y = 0$ at the p

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Let the line $L_{1}:x + 3 = 0$ intersect the lines $L_{2}:x - y = 0$ and $L_{3}:3x + y = 0$ at the points A and B , respectively. Let the bisector of the obtuse angle between the lines $L_{2}$ and $L_{3}$ intersect the line $L_{1}$ at the point C . Then ${BC}^{2}:{AC}^{2}$ is equal to :

Accepted Answer

Sol. [Key Concept: Find A, B and the obtuse-angle bisector point C; take the ratio on the transversal $L_{1}$.]$$\begin{matrix} \frac{{BC}^{2}}{{AC}^{2}} & \ = \left( \frac{OB}{OA} ight)^{2} = \frac{9 + 81}{9 + 9} \ & \ = 5 \end{matrix}$$

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