How to prove that triangle is rectangular My goal here is to prove that multiplication of segments of tangent line concluded between two straight lines (which are also tangents) are equal to $r^2$. And it's evident if given triangle has a corner of $ 90^\circ$. So i'm trying to find a way to prove that any triangle with a hypotenuse as a tangent line to this circle is rectangular. I think i could draw a median and make a circumcircle around this triangle, but i guess there is some other proof, not only to this, but also to any other tangent by different angle.

 A: Draw the line $CD$ through O perpendicular to the parallel lines: $C$ on the line going through $A$ and $D$ on the line going through $B$. Those parallel lines touch the circle at $C$ and $D$. Also, let $E$ be the point where line $AB$ touches the circle.
We have $\triangle OAC\cong\triangle OAE$ (because $OA=OA, OC=OE, \angle OCA=\angle OEA=90^\circ$) and similarly $\triangle OBD\cong\triangle OBE$.
Thus, $\angle AOE=\angle AOC$ and $\angle BOE=\angle BOD$, so $$\angle AOB=\angle AOE+\angle BOE=\frac{1}{2}(\angle AOC+\angle AOE+\angle BOE+\angle BOD)=\frac{1}{2}\cdot 180^\circ=90^\circ$$
A: Using coordinates, if $O=(0,0), B=(b,-1)$ and the circle has radius $1$, then $A=(1/b,1)$. The vectors $OA$ and $OB$ are clearly orthogonal.
Here is a roadmap.
Let $B=(b,-1)$. A line through $B$ is given by $y+1 = \alpha (x-b)$. Plug this into $x^2+y^2=1$ and get a quadratic equation in $x$. The line is tangent to the circle iff the discriminant is zero. This gives $\alpha=2b/(1-b^2)$. Plugging this and $y=1$ back into the line equation gives $A=(1/b,1)$.
