3. In the diagram, $O A B$ is a quarter circle, $O E$ and $C D$ are parallel, $B C$ and $E A$ are parallel, and $\angle B C D=$ $4 \times \angle O B C$.
What is the size of $\angle O B C$ ?

Quadrilateral $F C G E$ must be a parallelogram, as $O E$ is parallel to $C D$ and $B C$ is parallel to $E A$ (given in the question).

We have that $\angle B C D=\angle A E O=4 x^{\circ}$, as opposite angles in a parallelogram are equal and we are told in the question that $\angle B C D=4 \times \angle O B C$.

Triangle $O A E$ is isosceles as $O A=O E$ because both are radii of the quarter circle $O A B$, thus $\angle O A E=\angle A E O=$ $4 x^{\circ}$.

Also $\angle O C B=\angle O A E$, as $B C$ is parallel to $E A$ and hence they are corresponding angles.

As $O A B$ is a quarter circle then $\angle B O A=90^{\circ}$.
Considering the sum of the interior angles of the right-angled triangle $O C B$,
$$
180^{\circ}=90^{\circ}+4 x^{\circ}+x^{\circ}
$$

Finally, the previous equation simplifies to $90^{\circ}=5 x^{\circ}$ and so $\angle O B C=x^{\circ}=18^{\circ}$.
SolUtion
We begin by defining $F$ as the intersection of $O E$ and $B C$ and $G$ as the intersection of $C D$ and EA.

Let $\angle O B C=x^{\circ}$.