How can we prove that this triangle is Equilateral Triangle? This is a problem which was sent to me by a friend , but i couldn't solve it , in particular , i don't have ideas for that . 
I hope you can help  by hints or any thing .
Here is the problem in the image . 

 A: This is known as Morley's trisector theorem.
A: 
Morley's Theorem: Take any triangle and trisect its angles. For each side of the triangle, consider the point of intersection of the trisectors of the angles on that side that are closest to that side. The triangle which has these points as vertices is equilateral.

This is known as Morley's Triangle for Frank Morley who discovered it in 1899.
The proof presented here, which was posted on sci.math on 16 Mar 1996, starts with an equilateral triangle. Given any triangle, we construct a similar triangle that has this equilateral triangle at the intersections of its trisectors.  By similarity, the given triangle must have an equilateral triangle at the intersection of its trisectors.

Let the vertices of the equilateral triangle be $P$, $Q$, and $R$, and let the angles of the given triangle be $A$, $B$, and $C$.  Using the sides of $\triangle PQR$ as bases, erect the following isosceles triangles outside $\triangle PQR$:
$$
\triangle PSQ\quad\text{with}\quad\angle PSQ=\tfrac23A\\
\triangle QTR\quad\text{with}\quad\angle QTR=\tfrac23B\\
\triangle RUP\quad\text{with}\quad\angle RUP=\tfrac23C
$$
Draw the following circles:
$$
\text{$D_S$ centered at $S$ passing through $P$ and $Q$}\\
\text{$D_T$ centered at $T$ passing through $Q$ and $R$}\\
\text{$D_U$ centered at $U$ passing through $R$ and $P$}\\
$$
Add the following points:
$$
\text{$P_S$ and $Q_S$ on $D_S$ so that $\overline{P_SP}=\overline{PQ}=\overline{QQ_S}$}\\
\text{$Q_T$ and $R_T$ on $D_T$ so that $\overline{Q_TQ}=\overline{QR}=\overline{RR_T}$}\\
\text{$R_U$ and $P_U$ on $D_U$ so that $\overline{R_UR}=\overline{RP}=\overline{PP_U}$}\\
$$
Draw the following lines:
$$
\text{$L_P$ containing $P_U$ and $P_S$ (perpendicular to $\overline{PP_S}$ if $P_U=P_S$)}\\
\text{$L_Q$ containing $Q_S$ and $Q_T$ (perpendicular to $\overline{QQ_T}$ if $Q_S=Q_T$)}\\
\text{$L_R$ containing $R_T$ and $R_U$ (perpendicular to $\overline{RR_U}$ if $R_T=R_U$)}
$$
Add the following points:
$$
\text{$V_S$ at the intersection of $L_P$ and $L_Q$}\\
\text{$V_T$ at the intersection of $L_Q$ and $L_R$}\\
\text{$V_U$ at the intersection of $L_R$ and $L_P$}\\
$$
This construction is illustrated in the image above.
We will now show that $\triangle V_SV_TV_U$ has angles $A$, $B$, and $C$, and that $\triangle PQR$ is at the intersection of the trisectors of its angles.  We will implicitly use the fact that $A+B+C=\pi$.  Remember that $\triangle P_UPP_S$, $\triangle Q_SQQ_T$, and $\triangle R_TRR_U$ are isosceles.
By the construction above, it is easy to show that
$$
\left.\begin{align}
\angle P_SPQ&=\pi-\tfrac23A\\
\angle QPR&=\tfrac13\pi\\
\angle RPP_U&=\pi-\tfrac23C
\end{align}\right\}\angle P_UPP_S=\tfrac13(C+A-B)
$$
$$
\left.\begin{align}
\angle Q_TQR&=\pi-\tfrac23B\\
\angle RQP&=\tfrac13\pi\\
\angle PQQ_S&=\pi-\tfrac23A
\end{align}\right\}\angle Q_SQQ_T=\tfrac13(A+B-C)
$$
$$
\left.\begin{align}
\angle R_URP&=\pi-\tfrac23C\\
\angle PRQ&=\tfrac13\pi\\
\angle QRR_T&=\pi-\tfrac23B
\end{align}\right\}\angle R_TRR_U=\tfrac13(B+C-A)
$$
Consider the pentagon $PP_SV_SQ_SQ$ whose angles sum to $3\pi$:
$$
\begin{align}
\angle QPP_S&=\pi-\tfrac23A=\tfrac13(A+3B+3C)\\
\angle PP_SV_S&=\tfrac12(\pi+\angle P_UPP_S)=\tfrac13(2C+2A+B)\\
\angle V_SQ_SQ&=\tfrac12(\pi+\angle Q_SQQ_T)=\tfrac13(2A+2B+C)\\
\angle Q_SQP&=\pi-\tfrac23A=\tfrac13(A+3B+3C)\\
\text{total}&=2A+3B+3C=3\pi-A\\
\angle P_SV_SQ_S&=A
\end{align}
$$
Thus, $\angle P_SV_SQ_S=A$.  Because the central angle subtended by the chords $\overline{P_SP}$, $\overline{PQ}$, and $\overline{QQ_S}$ is $2A$, the locus of points from which the angle subtended by those chords is $A$ is the portion of $D_S$ not spanned by those chords.  Therefore, $V_S$ is on $D_S$ and it follows immediately that $\overline{PQ}$ subtends the middle third of $\angle P_SV_SQ_S$.
Similarly, $\angle Q_TV_TR_T=B$, $V_T$ is on $D_T$, and $\overline{QR}$ subtends
the middle third of $\angle Q_TV_TR_T$.
Similarly, $\angle R_UV_UP_U=C$, $V_U$ is on $D_U$, and $\overline{RP}$ subtends
the middle third of $\angle R_UV_UP_U$.
