About the ratio of the areas of a convex pentagon and the inner pentagon made by the five diagonals

Denote by $z_i$ $(0\leq i\leq 4)$ the vertices of the large pentagon $Z$ and by $w_i$ the vertices of the small pentagon $W$. WLOG we may assume $$z_0=(0,0), \quad z_2=(1,0),\quad z_3=(0,1)$$ and $$w_3=(a,0), \quad w_4=(b,0),\quad w_1=(0,d),\quad w_2=(0,c)$$ with $$0<a<b<1,\quad 0<c<d<1,\quad d<{c\over a}<{1\over b}\ .$$

One then obtains $$z_1=\left({ba(1-c)\over a-bc}, {c(a-b)\over a-bc}\right),\quad z_4=\left({a(c-d)\over c-ad}, {dc(1-a)\over c-ad}\right),\quad w_0=\left({b(1-d)\over 1-bd}, {d(1-b)\over 1-bd}\right)\ .$$ This means that the moduli space ${\cal M}$ of such configurations up to affinity has dimension $4$. The values $a=c={3-\sqrt{5}\over2}\doteq0.381966$ and $b=d={\sqrt{5}-1\over2}\doteq0.618034$ correspond to regular pentagons $Z$ and $W$.

Further computation then gives $$2\>{\rm area}(Z)=1+{(b-a)c\over a- bc}-{a(d-c)\over ad-c}\>,\quad 2\>{\rm area}(W)={bd(2-b-d)-ac(1-bd)\over 1-bd}\ ,\tag{1}$$ and we are interested in the quantity $$t(a,b,c,d):={{\rm area}(W)\over {\rm area}(Z)}\ .$$ For regular pentagons $Z$ and $W$ one has $t={7-3\sqrt{5}\over2}\doteq0.145898$.

The expressions $(1)$ are complicated enough to forbid calculating partial derivatives of $t$. On the other hand it is easy to see that $\inf_{\cal M} t(a,b,c,d)=0$. As for the maximum of $t$ I simulated $10^7$ random such configurations with Mathematica and obtained the following optimal quintuple $(a,b,c,d,t)$: $$(0.383542, 0.619248, 0.381882, 0.618277, 0.14589)\ .$$ This supports the conjecture that $t$ is maximal for regular pentagons.