How to justify this orthogonality?

Question

I am trying to justify the orthogonality of two vectors but I am stuck. First here is the figure :
I want to justify the orthogonality of vector CQ and vector PR. We know that :

• ABCD is a square.
• DR = AP
• P is on AB
• R is on AD

Here is a dot product : CQ.(AR-AP) (these are all vectors).
I develop it with : CQ.AR - CQ.AP.
I can say that CQ.AR = DR.AR and that CQ.AP = BP.AP.
It's here that I am stuck, I just don't know at all what to do ! BTW sorry, I don't know how to create the arrows for the vectors.

Answer 1

Let's take the side of the square to be equal $1$, and $|AP|=x$.

Then $P=(x,0), R=(0,1-x), Q=(x,1-x), C=(1,1)$, and the vectors are $$\overrightarrow{CQ} = (x-1,-x)$$ $$\overrightarrow{PR} = (-x,1-x)$$

Meaning $$\overrightarrow{CQ} \cdot \overrightarrow{PR} = (-x^2+x)+(-x+x^2) = 0$$

Answer 2

Let $i=\overrightarrow{AB}$, $j=\overrightarrow{AD}$. By assumption there is a real number $t\in(0,1)$ such that $\overrightarrow{AP}=ti$ and $\overrightarrow{AR}=(1-t)j$, So $\overrightarrow{AQ}=ti+(1-t)j$. Also, $AC=i+j$. Thus $$ \overrightarrow{PR}=\overrightarrow{AR}-\overrightarrow{AP}=-ti+(1-t)j,\quad \overrightarrow{CQ}=\overrightarrow{AQ}-\overrightarrow{AC}=(t-1)i-tj $$ So $$\overrightarrow{PR}\cdot\overrightarrow{CQ}=-t(t-1)-t(1-t)=0$$ Which is the desired conclusion.