Ans: A
Note that $\angle DAB$ and $\angle CBA$ are $90^\circ$. Therefore $AD$ and $BC$ have the greatest slope. Hence, $a$ and $b$ are the greatest angle.
Note that $\angle DAB$ and $\angle CBA$ are $90^\circ$. Therefore $AD$ and $BC$ have the greatest slope. Hence, $a$ and $b$ are the greatest angle.
It is given that $AG:GB=5:3$. Then we have $AG > GB$. Since $AE = BF$ and $AG > GB$, then we have $EG > FG$.
In $\Delta DEG$, $\tan d = \dfrac{DE}{EG}$. And in $\Delta CFG$, $\tan c = \dfrac{CF}{FG}$. Since $DE=CF$ and $EG > FG$, then we have
$\begin{array}{rcl}
\dfrac{DE}{EG} & < & \dfrac{CF}{FG} \\
\tan d & < & \tan c \\
d & < & c
\end{array}$
Hence, we have $a > c > d$.
