16) In the adjacent figure, $A \widehat{B} E=E \widehat{B} C$ and $A \widehat{C} E=E \widehat{C} D$. Knowing that $C \widehat{E} B=y$, then the angle $B \widehat{A} C$ is equal to
(A) $90^{\circ}-y$
(B) $90^{\circ}-\frac{y}{2}$
(C) $2 y$
(D) $180^{\circ}-2 y$
(E) $y$.

![](https://cdn.mathpix.com/cropped/2024_04_17_4c97eba1baa5e71bd15fg-2.jpg?height=250&width=392&top_left_y=1009&top_left_x=1072)
16) The answer is (C). Let $A \widehat{B} E=x$ and $A \widehat{C} E=z$, and call $O$ the intersection of $B E$ and $C A$. We will have that $B \widehat{O} C=O \widehat{C} E+C \widehat{E} O=z+y$ and $2 z=A \widehat{C} E=B \widehat{O} C+O \widehat{B} C=y+z+x$. From this, we get that $z-x=y$. Furthermore, $x+B \widehat{A} C=B \widehat{O} C=z+y$, from which $B \widehat{A} C=z+y-x=2 y$. Second Solution

Considering the triangle $B C E$ and its external angle $E \widehat{C} D$, we get $y=E \widehat{C} D-E \widehat{B} C$. Considering the triangle $A B C$ and its external angle $A \widehat{C} D$, we get $B \widehat{A} C=2 E \widehat{C} D-2 E \widehat{B} C=2 y$.