In the figure, ABDE is a trapezium and BCDE is a rhombus. AFD is a straight line. AE // BD, $\angle\text{ADB} = 21^\circ$ and $\angle\text{DBC} = 58^\circ$. Find $\angle\text{EFD}$. Ans: ___ $^\circ$

1. Find $\angle\text{EBD}$ using the properties of a rhombus.

In the rhombus $BCDE$, the diagonal $BD$ bisects the interior angle at $B$.
$\angle\text{EBD} = \angle\text{DBC} = 58^\circ$

2. Find $\angle\text{EFD}$ using the exterior angle theorem for $\triangle FBD$.

For $\triangle FBD$, $\angle\text{EFD}$ is an exterior angle.
$\angle\text{EFD} = \angle\text{FBD} + \angle\text{FDB}$
$\angle\text{EFD} = 58^\circ + 21^\circ = 79^\circ$