Please refrain from using algebraic equations (in the Cartesian system) to prove it. I was looking for some kind of geometric proof for it. Or using differentitiation of vectors. For using the definition of tangent, let the curve be described by $\vec r(t) $ and $\theta(t)$ for a parameter $t$. Then the tangent may be given by the direction of $\frac{d\vec r}{dt}$.
I got a start, let the origin be the center of the circle. Let $\vec m$ be the vector given by the radius of the circle pointing radially outwards ending at the point on the circle till where the string is bound to the circle, and $\vec n$ be the vector such that $\vec m+\vec n=\vec r$.
Now differentiating the vectors with respect to $t$, and using $\vec r=r.\hat r$. Now $\frac{dm}{dt}=0$.
So $m\frac{d\hat m}{dt}+n\frac{d\hat n}{dt}+\hat n\frac{dn}{dt}=$ tangent vector.
Now $|\vec n|=n=m\theta$, where $\theta$ is the angle rotated till now by the $\vec m$. noLet |$\frac{d\hat m}{dt}$| be constant. So $\frac{dn}{dt}=m|\frac{d\hat m}{dt}|$.
Now all that is left to show is that $\frac{d\hat m}{dt}$ is $\perp$ to $\vec m$. How do I do that?