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**earboth** Let $\displaystyle \alpha$ denote the angle $\displaystyle \angle(BOP)$

1. The position vector of A is (-r,0)

The position vector of B is (r, 0)

The position vector of P is $\displaystyle (r \cdot \cos(\alpha), r \cdot \sin(\alpha))$

2. Then

$\displaystyle \overrightarrow{AP}=(r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha))$ and

$\displaystyle \overrightarrow{BP}=(r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha))$

3. Now calculate

$\displaystyle \overrightarrow{AP} \cdot \overrightarrow{BP}= (r \cdot \cos(\alpha) + r, r \cdot \sin(\alpha)) \cdot (r \cdot \cos(\alpha) - r, r \cdot \sin(\alpha)) =$ $\displaystyle r^2\cos^2(\alpha)-r^2+r^2\sin^2(\alpha) = r^2-r^2 = 0$

And therefore the legs of a right triangle are perpendicular. (Thales theorem)

To the second question:

The cordinates of the squares vertices are O(0, 0), A(s, 0), B(s, s), C(0, s) where s means the side length.

Calculate the vectors of the diagonals and prove that the scalar product of these two vectors is zero.