In $\bigtriangleup OAB,\angle\mathrm{AOB}=90^\circ.$ Let C lie on segment $AB$ such that $\overrightarrow{OC}\;\perp\;\overrightarrow{AB}$.

Question

I am stuck at proving the following question.

In $\bigtriangleup OAB,\angle\mathrm{AOB}=90^\circ.$ Let C be the point on the segment $AB$ such that $\overrightarrow{OC}\;\perp\;\overrightarrow{AB}$.

Show that

$\displaystyle\frac{\overrightarrow{\left|CA\right|}\;}{\overrightarrow{\left|CB\right|}}=\frac{\displaystyle\overrightarrow{\left|OA\right|}^2}{\displaystyle\overrightarrow{\left|OB\right|}^2}$

EDIT: I got to $\overrightarrow{AC}\;=\lambda\;\overrightarrow{AB}$ and expressed $\overrightarrow{OC}\;$ in terms of $\overrightarrow{OA},\;\overrightarrow{OB}\;and\;\lambda$. I know $\overrightarrow{OC}\;.\;\overrightarrow{AB}\;=\;0$ but I do not know how to carry on

Answer 1

Since $\Delta AOC\sim\Delta OBC,$ we obtain: $$\frac{AC}{OC}=\frac{OC}{BC}=\frac{AO}{OB}.$$ Thus, $OC^2=AC\cdot BC$ and $$\frac{AO^2}{BO^2}=\frac{AC^2}{OC^2}=\frac{AC^2}{AC\cdot BC}=\frac{AC}{BC}.$$ Also, you can end your idea.

Indeed, $$\vec{OC}=\vec{OA}+\lambda\vec{AB}=\vec{OA}+\lambda\left(\vec{OB}-\vec{OA}\right)=(1-\lambda)\vec{OA}+\lambda\vec{OB}$$ and since $\vec{OC}\cdot\vec{AB}=0$ we obtain: $$\left((1-\lambda)\vec{OA}+\lambda\vec{OB}\right)\left(\vec{OB}-\vec{OA}\right)=0$$ or $$-(1-\lambda)OA^2+\lambda OB^2=0$$ or $$\lambda=\frac{OA^2}{OA^2+OB^2}=\frac{OA^2}{AB^2}.$$ Thus, $$AC=\lambda AB=\frac{OA^2}{AB},$$ $$BC=(1-\lambda)AB=\frac{OB^2}{AB}$$ and $$\frac{AC}{BC}=\frac{OA^2}{OB^2}.$$

Answer 2

Hint:

$$\triangle OAB\sim \triangle CAO\sim\triangle COB.$$

Can you see why? Can you see why that lets you conclude the result?