Inscribed triangle in a circle from perpendicular chords

Question

I'm trying to solve the following problem but have hit a wall.

Let's assume we have a circle. Said circle has a chord from point $A$ to point $B$. At one of the points where the chord crosses the circle, another chord is generated with a direction perpendicular to the first chord. So we get a new chord from $B$ to $C$.

Will then the line $\vec{CA}$ cross the center of the circle?

I have tried find it out by solving the equations of a line $y = mx + n$ and $x^2 + y^2 = r$. Solving that we would get the two points $(x_a,y_a)$ and $(x_b, y_b)$ from where we can generate the new vector. Then I solve the system again with the new line generated $y_2 = mx_2 + k$. And lastly I try to see that $(0,0)$ is a valid point for the line that goes from C to A. But the math is proving to be a bit too cumbersome for me and so far I haven't been able to solve it. Are there any cleverer ways to answer the question instead of trying to bruteforce it through solving the equations?

Any help is welcome. Thanks!!

I add a small sketch. To better illustrate the problem.

Answer 1

We will use the following picture. Suppose we have that points $A$, $B$, $C$ lie on circle $\omega$ with center $O$ and $\angle ABC = 90^{\circ}$. So, the triangle $ABC$ is right. Now, let $O^{'}$ is middle point of $AC$. Then $O^{'}A=O^{'}B=O^{'}C$ and $O^{'}$ is center of circumscribed circle of triangle $ABC$. But circumscribed circle of triangle $ABC$ is unique and coincides with $\omega$. We get that points $O$ and $O^{'}$ are coincide.

Answer 2

You can prove the inscribed angle theorem that inscribed angle is the half of central angle.

Now if central angle is $180^{\circ}$ (diameter that cross the center), the inscribed angle should be $90^{\circ}$ and vice versa (you could prove by finding angles' relationship)