Problem: Show that the following equation is true,$$ \begin{vmatrix} \frac{(a+b+c+d)(bcd+acd+abd+abc)}{4abcd} && \frac{a+b+c+d}{2} && \frac{bcd+acd+abd+abc}{2abcd} && 1 \\ \frac{( ab(c+d)-c d(a+b))(a+b-c-d)}{(a b-c d)^2} && \frac{a b(c+d)-c d(a+b)}{a b-c d} && \frac{a+b-c-d}{ab-cd} && 1 \\ \frac{(a d(c+b)-c b(a+d))(a+d-c-b)}{(a d-c b)^2} && \frac{a d(c+b)-c b(a+d)}{a d-c b} && \frac{a+d-c-b}{ad-cb} && 1 \\ \frac{(a c(b+d)-b d(a+c))(a+c-b-d)}{(a c-b d)^2} && \frac{a c(b+d)-b d(a+c)}{a c-b d} && \frac{a+c-b-d}{ac-bd} && 1 \notag \end{vmatrix}=0 $$ with $a,b,c,d \in \mathbb{C}$. What is the simplest way of establishing this result?
Context
In one of my answers, I used Mathematica to check a result. At first I was being lazy, but then I noticed that the equation was not easy to verify. It got me wondering whether it is possible to evaluate such an equation under contest conditions.
I was trying to show that the complex numbers $t, f, e, x$ are concyclic. I had defined them as follows ($e$ was obtained by swapping $b$'s and $d$'s in $f$ and $x$ was obtained by swapping $b$'s and $c$'s in $f$) $$t=\frac{a+b+c+d}{2}$$ $$f=\frac{a b(c+d)-c d(a+b)}{a b-c d}$$ $$ e=\frac{a d(c+b)-c b(a+d)}{a d-c b}$$ $$ x=\frac{a c(b+d)-b d(a+c)}{a c-b d} $$ with $a,b,c,d \;$ being complex numbers which lie on the unit circle. In my answer, I verified the following (concylic criterion) using Mathematica,
$$\frac{f-t}{e-t} \div \frac{f-x}{e-x} \; \in \; \mathbb{R} \tag{1}$$
but this equation is hard to verify by hand (though, it may still be easier than what I tried next).
There is another criterion to check whether the points lie on a circle, involving a $4 \; x\;4$ determinant. It can be shown that the four complex numbers $z_{1}, z_{2}, z_{3}, z_{4}$ lie on a generalized circle iff: $$\left|\begin{array}{cccc} z_{1} \overline{z_{1}} & z_{1} & \overline{z_{1}} & 1 \\ z_{2} \overline{z_{2}} & z_{2} & \overline{z_{2}} & 1 \\ z_{3} \overline{z_{3}} & z_{3} & \overline{z_{3}} & 1 \\ z_{4} \overline{z_{4}} & z_{4} & \overline{z_{4}} & 1 \end{array}\right|=0 $$
Therefore, we are trying to show that
$$ \begin{vmatrix} \dfrac{a+b+c+d}{2}\cdot\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}{2} & \dfrac{a+b+c+d}{2} & \dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}{2} && 1 \\ \dfrac{a b(c+d)-c d(a+b)}{a b-c d}\cdot\dfrac{\dfrac{1}{a} \dfrac{1}{b}(\dfrac{1}{c}+\dfrac{1}{d})-\dfrac{1}{c} \dfrac{1}{d}(\dfrac{1}{a}+\dfrac{1}{b})}{\dfrac{1}{a} \dfrac{1}{b}-\dfrac{1}{c} \dfrac{1}{d}} & \dfrac{a b(c+d)-c d(a+b)}{a b-c d} & \dfrac{\dfrac{1}{a} \dfrac{1}{b}(\dfrac{1}{c}+\dfrac{1}{d})-\dfrac{1}{c} \dfrac{1}{d}(\dfrac{1}{a}+\dfrac{1}{b})}{\dfrac{1}{a} \dfrac{1}{b}-\dfrac{1}{c} \dfrac{1}{d}} && 1 \\ \dfrac{a d(c+b)-c b(a+d)}{a d-c b}\cdot \dfrac{\dfrac{1}{a} \dfrac{1}{d}(\dfrac{1}{c}+\dfrac{1}{b})-\dfrac{1}{c} \dfrac{1}{b}(\dfrac{1}{a}+\dfrac{1}{d})}{\dfrac{1}{a} \dfrac{1}{d}-\dfrac{1}{c} \dfrac{1}{b}} & \dfrac{a d(c+b)-c b(a+d)}{a d-c b} & \dfrac{\dfrac{1}{a} \dfrac{1}{d}(\dfrac{1}{c}+\dfrac{1}{b})-\dfrac{1}{c} \dfrac{1}{b}(\dfrac{1}{a}+\dfrac{1}{d})}{\dfrac{1}{a} \dfrac{1}{d}-\dfrac{1}{c} \dfrac{1}{b}} && 1 \\ \dfrac{a c(b+d)-b d(a+c)}{a c-b d}\cdot\dfrac{\dfrac{1}{a} \dfrac{1}{c}(\dfrac{1}{b}+\dfrac{1}{d})-\dfrac{1}{b} \dfrac{1}{d}(\dfrac{1}{a}+\dfrac{1}{c})}{\dfrac{1}{a} \dfrac{1}{c}-\dfrac{1}{b} \dfrac{1}{d}} & \dfrac{a c(b+d)-b d(a+c)}{a c-b d} & \dfrac{\dfrac{1}{a} \dfrac{1}{c}(\dfrac{1}{b}+\dfrac{1}{d})-\dfrac{1}{b} \dfrac{1}{d}(\dfrac{1}{a}+\dfrac{1}{c})}{\dfrac{1}{a} \dfrac{1}{c}-\dfrac{1}{b} \dfrac{1}{d}} && 1 \notag \end{vmatrix} $$ is equal to zero. Simplifying the expressions: $$ \begin{vmatrix} \dfrac{(a+b+c+d)(bcd+acd+abd+abc)}{4abcd} & \dfrac{a+b+c+d}{2} & \dfrac{bcd+acd+abd+abc}{2abcd} & 1 \\ \dfrac{( ab(c+d)-c d(a+b))(a+b-c-d)}{(a b-c d)^2} & \dfrac{a b(c+d)-c d(a+b)}{a b-c d} & \dfrac{a+b-c-d}{ab-cd} & 1 \\ \dfrac{(a d(c+b)-c b(a+d))(a+d-c-b)}{(a d-c b)^2} & \dfrac{a d(c+b)-c b(a+d)}{a d-c b} & \dfrac{a+d-c-b}{ad-cb} & 1 \\ \dfrac{(a c(b+d)-b d(a+c))(a+c-b-d)}{(a c-b d)^2} & \dfrac{a c(b+d)-b d(a+c)}{a c-b d} & \dfrac{a+c-b-d}{ac-bd} & 1 \notag \end{vmatrix} $$ Since we are trying to show that the determinant is (a multiple of) zero, we are free to multiply rows and columns with arbitrary constants. I decided to get rid of the denominators
\begin{align*} \small \begin{vmatrix} (a+b+c+d)(bcd+acd+abd+abc) & (a+b+c+d)(2abcd) & {2(bcd+acd+abd+abc)} & 4abcd \\ ( ab(c+d)-c d(a+b))(a+b-c-d) & (a b(c+d)-c d(a+b))(a b-c d) & (a+b-c-d)(a b-c d) & (a b-c d)^2 \\ (a d(c+b)-c b(a+d))(a+d-c-b) & (a d(c+b)-c b(a+d))(a d-c b) & (a+d-c-b)(ad-cb) & (a d-c b)^2 \\ (a c(b+d)-b d(a+c))(a+c-b-d) & (a c(b+d)-b d(a+c))(a c-b d) & (a+c-b-d)(ac-bd) & (a c-b d)^2 \notag \end{vmatrix} \end{align*}
which resulted in another ugly-looking determinant. I hope a clever manipulation of the determinant exists (leading to a cyclic sum or something significantly simpler).
What is the easiest way to evaluate this determinant under contest conditions, or is my inital criterion, Equation $(1)$, more straightforward to verify?
Thank you in advance.
