Define a semi congruent number $\,n\,$ as a congruent number which represents the area of a triangle $\,\bigtriangleup ABC\,$ which has one leg in $\,\mathbb{Q}\setminus \mathbb{Z}\,$ and the other in $\mathbb{Z}.\,$ Are there integer solutions to the equation
$$a^2 + \left(\frac{c}{d}\right)^2= \left(\frac{e}{f}\right)^2$$
with $\,\gcd(c,d)=1=\gcd(e,f)\,$ such that $\,f\neq 1\neq d,\,$ and $\,d\vert a .\quad$ Earlier we discovered there is no non-trivial solutions unless $\,f=d.\quad$ If we restrict $\,d\vert a\,$ are there solutions? If so then the number $\quad n = \dfrac{1}{2} a\cdot \dfrac{c}{d}\quad$ is a semi-congruent number. Otherwise, there is no semi-congruent numbers.
[Pre edit, where OP didn't require that $\frac{1}{2} a \frac{c}{d}$ is an integer.]
With $d=f$, you want $(ad)^2 + (c)^2 = (e)^2$.
With $d \mid a, d \neq 1$, this means that $ad$ must not be square free.
So you're looking for a primitive pythagorean triplet where one of the legs has a square term. With that, we can divide by non-1 factors, to get a solution.
EG Starting with $4^2 + 3^2 = 5^2$, we get $2^2 +( \frac{3}{2} )^2 = (\frac{5}{2} ) ^2$.
Starting with $12^2 + 5^2 = 13^2$, we get $ 6^2 + ( \frac{5}{2} ) ^2 = ( \frac{13}{2} ) ^2$.
[Addressing the edit, where OP requires that $\frac{1}{2} a \frac{c}{d}$ is an integer]
A sufficient condition is $p^2 \mid ad$ where $ p > 2$ is a prime, and we set $ d = p$.
EG Starting with $45^2 + 28^2 = 53^2$, we get $ 15^2 + (\frac{28}{3} )^2 = (\frac{ 53}{3})^2$
Another sufficient condition is that $4^2 \mid ad$ and we set $ d = 2$.
EG Starting with $80^2 + 39^2 = 89^2$, we get $ 40^2 + (\frac{39}{2})^2 = ( \frac{89}{2} ) ^2$.