B2. In the figure, you see a triangle $A B C$. Angle $B$ is equal to 63 degrees. On side $A C$ lies a point $D$ and on side $B C$ lies a point $E$. It is given that $B, A$, and $D$ lie on a circle with center $E$. Also, points $E$ and $C$ lie on a circle with center $D$.

How large is angle $C$?

(Note: the figure below is not to scale.)

![](https://cdn.mathpix.com/cropped/2024_04_17_b17c3267eb783a5c7d68g-1.jpg?height=380&width=546&top_left_y=1438&top_left_x=1326)
2. 18 degrees

Let the angle we are looking for, angle $C$, be $x$. Since $D$ is the center of a circle passing through $C$ and $E$, $C$ and $E$ are equidistant from $D$. Therefore, triangle $CDE$ is isosceles with vertex angle $D$. We see that $\angle DEC = x$ and $\angle CDE = 180^{\circ} - 2x$, because the sum of the angles in a triangle is $180^{\circ}$.

We now see that $\angle ADE = 2x$, because this angle is supplementary to $\angle CDE$. Triangle $AED$ is also isosceles, with

![](https://cdn.mathpix.com/cropped/2024_04_17_51fa63b20784abd70d25g-1.jpg?height=391&width=537&top_left_y=1209&top_left_x=1291)
vertex angle $E$. Thus, $\angle DAE = 2x$. Furthermore, triangle $ABE$ is also isosceles, with vertex angle $E$. We see that $\angle EAB = \angle ABE = 63^{\circ}$.

The sum of the angles in triangle $ABC$ is 180 degrees. Therefore, $63^{\circ} + (63^{\circ} + 2x) + x = 180^{\circ}$. This gives us $3x = 180^{\circ} - 126^{\circ} = 54^{\circ}$. The angle we are looking for is thus $\frac{1}{3} \cdot 54^{\circ} = 18^{\circ}$.