Vector proof to show the line connecting two points on a triangle is parallel to a side.

Question

If $X$ and $Y$ are points on sides $AB$ and $AC$ of a triangle $ABC$ and $\dfrac{AX}{AB}=\dfrac{AY}{AC}$, then $XY\parallel BC$.

I'm supposed to prove this using vectors, but we haven't done too much of this yet, and I'm at a loss.

Answer 1

Hint: Assume that $$\dfrac {AX}{AB} = \dfrac{AY}{AC} = \lambda.$$

It is easy to prove that $$\vec {XY} = \vec {AY} - \vec{AX}.$$

Similarly, $$\vec{BC} = \vec{AC} - \vec{AB}.$$

However, $$\vec{AX} = \lambda \vec{AB} \text{ and } \vec{AY} = \lambda \vec{AC} \quad \text{(why?)}$$

Combine all the facts above and you will reach the result.

Answer 2

let $t=\frac{AX}{AB}=\frac{AY}{AC}$

then $\vec{XY} = \vec{XA}+\vec{AY} = t(-\vec{AB}+\vec{AC}) $

and $\vec{BC} = -\vec{AB}+\vec{AC} $

so the two vectors are scalar multiples of each other, hence they are parallel.