<Qua-2025-791>, hint

Consider a isosceles trapezoid ABCD  AB//CD BC=AD, AB=a BC=c, CD=b

suppose a line through B, parallel to AC meet line DC and line DA at E', E".

Then the areas of AE'D,CE"D  are equal to the area of ABCD. (CE'=a CD=b)

Since DA'=(a+b)/2, A'B'=(b+a)/2-(b-a)/2=a. 

Suppose  line A'C'  meet line BA at C". Since AC' :C'D=(b-a)/2 : (b+a)/2

AC"=A'D x AC'/C'D=(b+a)/2 x (b-a)/(b+a)=(b-a)/2. 

line DA' meet line AB at D", BD"=AC". Hence C"D"=a+2{(b-a)/2}=b.

Finally A'B'C"D"  and ABCD are congruent  and symmetric wrt midline  ABCD.