In a triangle $\mathrm{ABC}$, $\mathrm{X}$ is given over $\mathrm{AB}$ such that $\|\overrightarrow{A X}\|=2\|\overrightarrow{X B}\| $ and given $\mathrm{Y}$ over $\mathrm{BC}$ such that $\|\overrightarrow{B Y}\|=$ $3\|\overrightarrow{Y C}\|$. Show that the lines CX and AY intersect.
Attempt: I took $\overrightarrow{AB}=3\overrightarrow{a}$, $\overrightarrow{BC} = 4 \overrightarrow{b}$. So I thought I'd prove that if $\overrightarrow{AX}$ and $\overrightarrow{YC}$ have the same direction, then $\overrightarrow{AY}$ and $\overrightarrow{XC}$ intersect. But I don't know how to prove it.... that was my expressions
I) $$\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC}= 3 \overrightarrow{a} + 4 \overrightarrow{b}$$ II) $$\overrightarrow{AC}=\overrightarrow{AY}+\overrightarrow{YC}$$ $$3\overrightarrow{a}+4\overrightarrow{b} = \overrightarrow{AY}+\overrightarrow{b}$$ $$\overrightarrow{AY}= 3(\overrightarrow{a}+\overrightarrow{b})$$
III) $$\overrightarrow{AC}=\overrightarrow{AX}+\overrightarrow{XC}$$ $$\overrightarrow{XC}= \overrightarrow{a}+\overrightarrow{4b}$$
I'd like to check if I'm correct:
Let $\overrightarrow{AB}=3\overrightarrow{a}$, $\overrightarrow{BC} = 4 \overrightarrow{b}$. We have
I) $$\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC}= 3 \overrightarrow{a} + 4 \overrightarrow{b}$$ II) $$\overrightarrow{AC}=\overrightarrow{AY}+\overrightarrow{YC}$$ $$3\overrightarrow{a}+4\overrightarrow{b} = \overrightarrow{AY}+\overrightarrow{b}$$ $$\overrightarrow{AY}= 3(\overrightarrow{a}+\overrightarrow{b})$$
III) $$\overrightarrow{AC}=\overrightarrow{AX}+\overrightarrow{XC}$$ $$\overrightarrow{XC}= \overrightarrow{a}+\overrightarrow{4b}$$
We will prove that $\nexists k$, such that $\overrightarrow{CX} = k \overrightarrow{AY}$, that is, that $AY$ and $CX$ are not parallel: $$k \overrightarrow{AY}+\overrightarrow{CX}= \overrightarrow{a}+4\overrightarrow{b}=k(3\overrightarrow{a}+3\overrightarrow{b}) \leftrightarrow (1- 3k )\overrightarrow{a}+(4-3k)\overrightarrow{b}=\overrightarrow{0}$$ See that since $\overrightarrow{a}$ and $\overrightarrow{b}$ represent sides of a triangle that are not collinear, so $\overrightarrow{a}$ and $\overrightarrow{b}$ are linearly independent and thus, by definition, the only way to get a vector $\overrightarrow{0}$ from linear combination of $\overrightarrow{a}$ and $\overrightarrow{b}$ is if $1-3k=0$ and $4-3k=0$. However, as the equations are contradictory, there is no scalar that makes $\overrightarrow{CX}$ and $\overrightarrow{AY}$ parallel.