Solution:
Answer: We can get arbitrarily close to (but not achieve) $\dfrac{3 \sqrt{3}}{4}$ (approx $1.3$) by:

To prove the first part, consider the diagram below. Take $AB = AC = 1$ and $\angle BAC = 2\theta$. Take $DE = DF = 1$ and take the points of intersection $X$ and $Y$ such that $AX = DX = AY = DY = 2/3$. It is easy to check that the area of the hexagon is $\sin 2\theta$. So by taking $\theta$ in the interval $(0, \pi/4]$ we can get any area $0 < k \leq 1$.

It is easy to check that there are six possible configurations for the unit diagonals, as shown in the diagram below.

Consider case 1.

The area of the hexagon is area $AEDC$ + area $AFE$ + area $BAC$. The part of the segment $BF$ that lies inside $AEDC$ is wasted. The rest goes to provide height for the triangles on bases $AE$ and $AC$. So area $AFE$ + area $BAC$ can be maximised by taking $F$ close to $A$ and $\angle BAC$ as close to a right angle as possible, so that the height of the triangle $BAC$ (on the base $AC$) is as large as possible. We can then get arbitrarily close to the area of:

We obviously make $AEB$ a straight line. Now area $ADE$ + area $ADC$ = area $ACE$ + area $CDE$. So if we regard every point except $D$ as fixed, then we maximise the area by taking $\angle EAD = \angle CAD$, so that $D$ is the maximum distance from $CE$. Thus a maximal configuration must have $\angle AED = \angle CAD$. Similarly, it must have $\angle CAD = \angle CAB$, so all three angles must be equal. That disposes of case 1.

In cases 2 and 6 we find by a similar (but more tedious argument) the same maximum, although in one case we have to use the argument at the end for the final optimisation. In the other cases the maximum is smaller.

4
5

6

However, all these details would take an already long solution way over length. Does anyone have a better approach?

No. 6 (second case) can be made arbitrarily close to the figure below (with $AB = AC = BD = 1$). To optimise it, suppose $\angle ACB = \theta$. Area $ABDC =$ area $ABC$ + area $BCD$. If we fix $\theta$, then $BC$ is fixed, so to maximise area $BCD$ we must take $\angle CBD = 90^\circ$. But $\theta$ cannot be optimal unless also $\angle CAD = 90^\circ$. We have $BA = BD$ and hence $\angle BAD = \angle BDA = 45^\circ - \theta/2$. Hence

$$90^\circ = \angle CAD = \angle BAC - \angle BAD = (180^\circ - 2\theta) - (45^\circ - \theta/2)$$

Hence $\theta = 30^\circ$. So $\angle ACD = \angle BDC = 60^\circ$ and $\angle CAB = \angle ABD = 120^\circ$. It is easy to check that this has area $\dfrac{3\sqrt{3}}{4}$.