P.S:I am looking for a hint and not the whole solution.
BdMO 2012 nationals secondary:
The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The
right angle is at $A$, so that the opposite arc BC is a semicircle while arc $AB$ and arc $AC$ are
supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the
midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$. More precisely, the
point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where the tangent at $D$
intersects the extended lines $AB$ and $AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$. Prove that triangle
$DEF$ is equilateral.
I have tried angle chasing,POP,forming pedal triangles,looking for cyclic quadrilaterals etc..A prod in the correct direction will be appreciated.
NOTE:If possible,please use an elementary approach to the problem.This contest usually emphasizes on Euclidean geometry so there must be a solution using Euclidean geometry.
Best Answer
Per light of @Blue's last comment, here's a (very brief) solution:
Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$.
Similarly, $F$ trisects the small arc $AB$ and $D$ trisects the large arc $AC$ (and $AB$).
That would show $\triangle DEF$ is equilateral.