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I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?

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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 $OP\lt OQ$.

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Suppose $\Gamma$ is a circle centered at $O$. Let $\ell$ be a line tangent to $\Gamma$ at a point $A$. Suppose the line from $O$ perpendicular to $\ell$ meets $\ell$ at a point $B$. If $B\neq A$, then there exists a point $C$ on $\ell$ on the other side of $B$ from $A$ such that $AB\cong BC$. (This follows from Hilbert's first axiom of congruence for line segments.)

By the side-angle-side theorem, $\triangle OBA\cong\triangle OBC$, and thus $OA\cong OC$. Thus $C\in\Gamma$. But $C\neq A$, a contradiction, since the tangent line $\ell$ can only meet $\Gamma$ at one point by definition of tangency. Thus $B=A$. Thus $\ell\perp OA$.

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Basically, since the tangent is perpendicular to the radius drawn to the point of contact and the chord is parallel to the tangent, thus the chord is perpendicular to the radius drawn to the point of contact.

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