An equilateral triangle formed using points of tangency 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.
 A: 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.
A: Coordinatize the triangle on the unit circle via
$$A = (\cos\alpha, \sin\alpha)\qquad B = (-1,0) \qquad C = (1,0)$$ 
Let $T$ be a point of tangency representing generically one of $D$, $E$, $F$.$$T = (\cos\theta,\sin\theta)$$
Define $T_B$ and $T_C$ as points on the tangent line, at distance $t$ from $T$; we can express these as $T \pm t\;T^\perp$, where $T^\perp$ is a unit vector orthogonal to $T$:
$$T_B = T+t\;(\sin\theta,-\cos\theta) \qquad T_C = T-t\;(\sin\theta,-\cos\theta)$$
Forcing $T_B$ to be on line $\overleftrightarrow{AB}$, and $T_C$ to be on line $\overleftrightarrow{AC}$, gives two conditions on the parameter $t$. Eliminating $t$ leaves an equation involving only $\theta$ (and $\alpha$), which has exactly three distinct solutions for $\theta$ equally-spaced about the circle; these are $D$, $E$, $F$.

The algebra is much easier if you coordinatize in the complex plane:
$$A = e^{i\alpha}\qquad B = -1 \qquad C = 1$$
Note that, given $T = e^{i\theta}$, we can write $T^\perp$ as $i\;T$, so that
$$T_B =  e^{i\theta} \left( 1 + it \right) \qquad T_C =  e^{i\theta} \left( 1 - it \right)$$ 
As before, imposing collinearity on $A$, $B$, $T_B$, and on $A$, $C$, $T_C$, gives equations in $t$, namely ...
$$\begin{align}
\frac{B - A}{T_B - A} = \frac{\overline{B} - \overline{A}}{\overline{T_B} - \overline{A}} &\quad\to\quad it ( e^{2i\theta} + e^{i\alpha}) = -(e^{i\theta} + 1) (e^{i\theta} - e^{i\alpha}) \\[4pt]
\frac{C - A}{T_C - A} = \frac{\overline{C} - \overline{A}}{\overline{T_C} - \overline{A}} &\quad\to\quad it (e^{2i\theta} - e^{i\alpha}) = \phantom{-}(e^{i\theta}- 1) (e^{i\theta} - e^{i\alpha})
\end{align}$$
Eliminating $t$ here gives an uncomplicated equation 
$$\left(\;e^{i\theta}-e^{i\alpha}\;\right)\left(\;e^{3i\theta}-e^{i\alpha}\;\right)=0$$
with an extraneous root $\theta = \alpha$, and with three symmetrically-arranged solutions $\theta = \frac{1}{3}(\alpha + 2k\pi)$ for $k=0,1,2$, indicating that $D$, $E$, $F$ are vertices of an equilateral triangle.

The results of the algebraic approach make the geometry of the situation clear: $E$ and $F$ trisect respective arcs $\stackrel{\frown}{AC}$ and $\stackrel{\frown}{AB}$. ($D$'s position is a bit harder to describe.) @QuangHoang posted a demonstration of this fact while I was working on my own; for completeness, here's my analysis:
 
Since $\angle PAQ = 90^\circ$, point $A$ lies on the circle about $E$ with diameter $PQ$.
$$\frac{1}{2}\stackrel{\frown}{AE} \;=\; \angle AEP \;=\; 2 \angle EAQ \;=\; \stackrel{\frown}{EC}$$
Thus, $E$ trisects $\stackrel{\frown}{AC}$. Likewise, $F$ trisects $\stackrel{\frown}{AB}$. As @QuangHoang notes, point $D$'s location isn't at all difficult to describe: it simultaneously trisects major arcs $\stackrel{\frown}{AB}$ and $\stackrel{\frown}{AC}$.
