How to prove that the tangent to a circle is perpendicular to the radius drawn to the point of contact? I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?
Let $O$ be the centre of the circle, let $\ell$ be a tangent line, and let $P$ be the point of tangency.  Suppose that $OP$ is not perpendicular to $\ell$. Draw the line through $O$ which is perpendicular to $\ell$. Then this line meets $\ell$ at a point $Q\ne P$.

Note that $Q$ is outside the circle. Now consider the triangle $OQP$. This is right-angled at $Q$.  So $OP$ is the hypotenuse of this triangle, and is therefore bigger than $OQ$. But this is impossible: since $Q$ is outside the circle, we must have $OP\lt OQ$.
Extend the radius to a line $L$ and use reflection symmetry with respect to this line.