Does Euclid's bisection of an angle fail for $\frac \pi 3$? Does Euclid's bisection of an angle fail for angles of $\frac \pi 3$ (that is, of equilateral triangles)?
In Book I Prop. 9 Euclid shows how to bisect an angle by constructing an equilateral triangle across it, and then connecting the angle's vertex to a point on the triangle.  A modern version of this is given on ProofWiki.  But these constructions seem to assume the point on the equilateral triangle will be distinct from the vertex of the angle, which is not necessarily true if the angle is $\frac \pi 3$.
Of course, the repair to the construction is easy: just construct the equilateral triangle on the other side of the line.  But the fact that such a triangle exists is never shown by Euclid (I.1) or by the modernized proofs on ProofWiki.
Updates
To address the comments:

In the diagram, we wish to bisect angle $BAC$, and we do so by constructing an equilateral triangle $FED$.  The problem is that if the angle is $\frac \pi 3$, we may get point $F$ is point $A$. The construction only works when $F$ is distinct from $A$.
It is well known that there are many omissions in Elements.  However, to my knowledge, these are always cases where Euclid assumes something without (explicitly) proving it; but, in each case, the fact is indeed provable (at least with a sufficient set of axioms).  This case is different, because it is indeed quite possible for $A = F$, in which case a different construction is needed.  Are there any other known cases where Elements does something like that?
 A: One of the known defects in Euclid's axiom system is the missing discussion of "betweenness". That's important for triples of points on a line. In this case the issue is the analogous one of "sidedness" that can distinguish between sides of a line.
See
Hilbert’s Axioms for Euclidean Plane Geometry
and
https://faculty.etsu.edu/gardnerr/Geometry/notes-Wylie/Geometry-Wylie-2-2.pdf
A: I believe the following proof is accessible via Euclid's axioms.
Claim: Let two circles centered at points $P$ and $Q$ intersect at a point A lying off $\overline{PQ}$. Then the circles also intersect at a point $B$ distinct from $A$.
Proof: Construct the perpendicular to $\overline{PQ}$ through $A$ which intersects $\overline{PQ}$ at $M$. Extend $\overline {AM}$ past $M$ to point $B$ such that $\overline{MB}$ is congruent with $\overline{MA}$. Then $\triangle PMA$ is congruent with $\triangle PMB$ via SAS, so corresponding sides $\overline{PA}$ and $\overline{PB}$ are also congruent proving that $B$ lies on $\bigcirc P$. Similarly $\triangle QMA$ is congruent with $\triangle QMB$ leading to $B$ lying on $\bigcirc Q$.
If this proof is valid, it implies that we can choose either of two points of intersection when we construct the arcs, and at least one is perforce distinct from the vertex of the angle.
