Orthogonal chords of compact sets For any compact set on a plane say C does there always exist a chord in C     such that its end points are orthogonal to the boundary of C (assumed smooth)
 A: Consider two points $p,q$ on the boundary at a maximal distance from each other. If the chord $pq$ is not orthogonal to the boundary, say at $p$, then one could move $p$ along the curve in one of the two directions, obtaining a new point $p'$ so that $d(p',q)>d(p,q)$. 
In a comment, Neal has asked what guarantees the existence of points $p,q$ on the boundary at maximal distance from each other. First notation: Let $A$ be the set in question, whose boundary is say the closed smooth curve $\gamma$. Since $A$ is compact it has a diameter, and so there are points $p_1,q_1$ for which $d(p_1,q_1)=d$ where $d$ is the diameter of $A$. For this result see
diameter on a compact metric space
Now we claim that in fact $p_1$ lies on the boundary of $A$. For if not, it lies in the interior of $A$ and so lies in some disc $D$ contained in $A$. But then we could extend the segment $p_1p_2$ a bit into $D$, increasing its length, against $d(p_1,q_1)=d$. Similarly we can show that $q_1$ must lie on the boundary. Conclusion: with $p=p_1,q=q_1$ we have two points on the boundary at maximal distance from each other, and in addition achieving the diameter of the whole set $A$.
A: The maximum distance argument is the essence of the matter (see answer by coffeemath) and I want to isolate here the minimal assumption needed for it to work, without imposing regularity conditions or topological restrictions on the boundary.  The compact set could be the Hawaiian Earring space, or an earring with uncountably many circles, or the boundary could be a figure $8$, but the proof continues to work if... 

  
*
  
*(infinitely near rays extend to infinitely near lines) for every boundary point $p$ of the boundary, if there is a sequence of points of the boundary $q_n$ converging to $p$ and asymptotic to a vector $v$ based at that point (so that the angle between $v$ and the vector from $p$ to $q_n$ goes
  to zero) then there are points of the boundary $r_n$ converging to $p$ and asymptotic to the vector opposite to $v$, AND  
  
*(all points have infinitely near tangent-like lines) Every point on the boundary has such a sequence for at least one nonzero vector $v$.  
  

This is true if the boundary is a union of differentiable curves with nonzero velocity vector (locally near any point, the curves are "branches", each one enclosing the point in an open smooth arc), which covers most cases.  The proof works in any number of dimensions, or in a Riemannian manifold if you understand $v$ to live in the tangent space at $p$.
