From the figure (same as q5219-q5220), find \( \angle EFG \) using angle sum of triangle EFG and the straight line BCEGH.<br><br>
Then express as a fraction: \( \dfrac{\angle EFG}{\angle FEG} \). Simplify to lowest terms.<br><br>
\( \dfrac{\angle EFG}{\angle FEG} = \dfrac{1}{2} \).<br><br>
Final Answer: A = 1, B = 2