In the figure below, EFGH is a rhombus and FGJK is a trapezium. EH and KF are straight lines. $\angle KLH = 55^\circ$ and $\angle HEF = 110^\circ$

1. Find the angle ELF using vertically opposite angles or angles on a straight line.

Since $K, L, F$ are on a straight line, $\angle ELF$ and $\angle KLH$ are vertically opposite angles. Thus,
$\angle ELF = \angle KLH = 55^\circ$

2. Find the angle EFL in triangle EFL.

In triangle $EFL$, the sum of angles in a triangle is $180^\circ$. Given $\angle HEF = 110^\circ$ (which is $\angle LEF$ since $L$ lies on $EH$):
$\angle EFL = 180^\circ - 110^\circ - 55^\circ = 15^\circ$

3. Find the diagonal angle EFH in the rhombus.

In rhombus $EFGH$, the diagonal $FH$ bisects the interior angle $\angle HEF$ (or we can use the isosceles triangle $EFH$ where $EF = EH$):
$\angle EFH = \frac{180^\circ - 110^\circ}{2} = 35^\circ$

4. Calculate angle KFH.

Since $L$ lies on the line segment $KF$, the angle $\angle KFH$ is the same as $\angle LFH$. Therefore:
$\angle KFH = \angle EFH - \angle EFL = 35^\circ - 15^\circ = 20^\circ$