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…

The intersection point of $L_1: x = -3$L1​:x=−3 and $L_2: x - y = 0$L2​:x−y=0 is $A(-3, -3)$A(−3,−3).

The intersection point of $L_1: x = -3$L1​:x=−3 and $L_3: 3x + y = 0$L3​:3x+y=0 is $B(-3, 9)$B(−3,9).

The lines $L_2$L2​ and $L_3$L3​ intersect at the origin $O(0, 0)$O(0,0).

The distances from the origin to $A$A and $B$B are:

$$OA = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}$$OA=(−3)2+(−3)2​=32​

$$OB = \sqrt{(-3)^2 + 9^2} = 3\sqrt{10}$$OB=(−3)2+92​=310​

The angle $\angle AOB$∠AOB between the segments $OA$OA and $OB$OB can be determined using the dot product of vectors $\overrightarrow{OA}$OA and $\overrightarrow{OB}$OB:

$$\overrightarrow{OA} \cdot \overrightarrow{OB} = (-3)(-3) + (-3)(9) = 9 - 27 = -18$$OA⋅OB=(−3)(−3)+(−3)(9)=9−27=−18

Since the dot product is negative, $\cos(\angle AOB) < 0$cos(∠AOB)<0, which means $\angle AOB$∠AOB is an obtuse angle. Therefore, the internal bisector of $\angle AOB$∠AOB is the bisector of the obtuse angle between the lines $L_2$L2​ and $L_3$L3​.

According to the internal angle bisector theorem in $\triangle AOB$△AOB, the bisector of $\angle AOB$∠AOB divides the opposite side $AB$AB at point $C$C in the ratio of the adjacent sides:

$$\dfrac{AC}{BC} = \dfrac{OA}{OB} = \dfrac{3\sqrt{2}}{3\sqrt{10}} = \dfrac{1}{\sqrt{5}}$$BCAC​=OBOA​=310​32​​=5​1​

Squaring both sides yields:

$$\dfrac{AC^2}{BC^2} = \dfrac{1}{5} \Rightarrow \dfrac{BC^2}{AC^2} = 5$$BC2AC2​=51​⇒AC2BC2​=5

Thus, $BC^2 : AC^2 = 5 : 1$BC2:AC2=5:1.

Answer: $5:1$5:1