In the diagram, point $E$ lies on line segment $A B$, and triangles $A E D$ and $B E C$ are isosceles. Also, $\angle D E C$ is twice $\angle A D E$. What is the size of $\angle E B C$ ?
(A) $75^{\circ}$
(B) $80^{\circ}$
(C) $60^{\circ}$
(D) $55^{\circ}$
(E) $45^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-105.jpg?height=361&width=512&top_left_y=1782&top_left_x=1251)
Since $\triangle A D E$ is isosceles, then $\angle A E D=\angle E A D=70^{\circ}$.

Since the angles in $\triangle A D E$ add to $180^{\circ}$, then $\angle A D E=180^{\circ}-2\left(70^{\circ}\right)=40^{\circ}$.

Since $\angle D E C=2(\angle A D E)$, then $\angle D E C=2\left(40^{\circ}\right)=80^{\circ}$.

Since $A E B$ is a straight line, then $\angle C E B=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$.

Since $\triangle E B C$ is isosceles, then $\angle E C B=\angle E B C$.

Thus, in $\triangle E B C, 30^{\circ}+2(\angle E B C)=180^{\circ}$ or $2(\angle E B C)=150^{\circ}$ or $\angle E B C=75^{\circ}$.

ANSWER: (A)