Does there exist a midsquare with natural sides and diagonal?

Question

A midsquare is an orthodiagonal and equidiagonal quadrilateral.

Are there naturals $a,b,c,d,e$ such that the quadrilateral with sides $a,b,c,d$ and diagonal $e$ is a midsquare ?

$e^2$ is a solution for $$2e^4-2(a^2+c^2)e^2+(a^2-b^2)^2+(b^2-c^2)^2=0$$ Let $D=4b^2d^2-(a^2-c^2)^2$
Then $e^2=\dfrac{a^2+c^2+\sqrt{D}}2$

p=1 
q=5000
import math
for a in range (p,q): # a = AB 
    for b in range(a,q): # b = BC
        for c in range (p,q): # c = CD
            if a*a+c*c > b*b:
               d = (a*a+c*c-b*b)**.5 # d = DA
               if int(d)==d:
                  d=int(d)
                  D = 4*b*b*d*d-(a*a-c*c)*(a*a-c*c) 
                  if D>=0:
                     if int(D**.5)==D**.5:
                        e = (.5*(a*a+c*c+D**.5))**.5
                        if int(e)==e: 
                           print(a+b+c+d+e+e,a,b,c,d,e,e)

Some results with $\sqrt{D}\in\mathbb{N}$
$$\begin{array}{|c|c|c|c|}\hline \text{perimeter-plus } a+b+c+d+2e & \sqrt{D} & e^2&e\\ \hline 658.488\approx35+101+149+115+2\times 129.244& 9982 & 16704&24\sqrt{29}\\ \hline 1183.63\approx41+181+289+229+2\times 221.815& 13202 & 49202& \sqrt{2\times73\times337}\\ \hline 3738.168\approx221+481+899+791+2\times673.084& 49042 & 453042& 3\sqrt{2\times25169}\\ \hline 6105.496\approx299+925+1405+1099+2\times1188.748&762818 & 1413122& \sqrt{2\times706561}\\ \hline 7611.367\approx85+1339+1939+1405+2\times1421.684&275422& 2021184& 264\sqrt{29}\\ \hline 11665.227\approx845+1513+2525+2191+2\times2295.614& 3450034 & 5269842& 3\sqrt{2\times137\times2137}\\ \hline 11992.528\approx1015+1241+2759+2665+2\times2156.264& 656642 & 4649474& \sqrt{2\times661\times3517}\\ \hline 17069.827\approx709+2759+3995+2975+2\times3315.913 &5527858& 10995282 & 3\sqrt{2\times610849}\\ \hline 21602.357\approx1681+2381+4889+4589+2\times4031.178&5772718 &16250400& 60\sqrt{2\times37\times61}\\ \hline 37947.648\approx1321+6329+9049+6601+2\times 7323.824 &23647358 & 53638400&1360\sqrt{29}\\ \hline 44392.996\approx2509+7081+9901+7361+2\times8770.498&49518382&76921632&12\sqrt{2\times89\times3001}\\ \hline \end{array}$$

Two special cases

1. The isosceles trapezoid midsquare $a,b,c,b,e,e$.
   It's easy to show that with $a,b,c,e$ integers, it doesn't exist: a nice geometric expression for $e$ as a function of $a$ and $c$.
2. The kite midsquare $a,a,c,c,e,e,~a<c$.
   I don't know whether it exists or not.
   For it, I can't even find an integer $\sqrt{D}=\sqrt{4a^2c^2-(a^2-c^2)^2}$.
   20/11/23 Let $g$ be the distance between the midpoints of the diagonals (see Euler's quadrilateral theorem), we can show that $$a^2+2eg=c^2$$ So $g$ is a rational number, we can look for $e$ and $g$ naturals such that $$(e-g)^2+g^2=2a^2$$ $$(e+g)^2+g^2=2c^2$$ Example of an "almost solution" $$(1192-1043)^2+1043^2=1~110~050=2\times745^2~~~~~~~~~~$$ $$(1192+1043)^2+1043^2=6~083~074=2\times1744^2~\boxed{+2}$$ I still don't find any solution.