Let $ABC$ be a triangle and $P$ a point on the side $BC$. Let $S_1$ be the circumference with center $B$ and radius $BP$ that cuts the side $AB$ at $D$ such that $D$ lies between $A$ and $B$. Let $S_2$ be the circumference with center $C$ and radius $CP$ that cuts the side $AC$ at $E$ such that $E$ lies between $A$ and $C$. Line $AP$ cuts $S_1$ and $S_2$ at $X$ and $Y$ different from $P$, respectively. We call $T$ the point of intersection of $DX$ and $EY$. Prove that $\angle BAC+ 2 \angle DTE=180$