We are given that $A\left(\overrightarrow{a}\right), B\left(\overrightarrow{b}\right), C\left(\overrightarrow{c}\right) \& D\left(\overrightarrow{d}\right)$ are four concyclic points such that

$x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}+t\overrightarrow{d}=\overrightarrow{0} ...\left(\text{i}\right)$

$x+y+z+t=0 ...\left(\text{ii}\right)$

Now, we know that angle subtended by chord on same side of circle are equal therefore 

$$\begin{gathered} \angle A=\angle C \\ \angle B=\angle D \\ \angle P=\angle P \end{gathered}$$

Hence, by $AAA$

$△APD~∆CPB$

So,

$\frac{AP}{CP}=\frac{PD}{PB}$

$\Rightarrow \frac{r-\overrightarrow{a}}{r-\overrightarrow{c}}=\frac{r-\overrightarrow{d}}{r-\overrightarrow{b}} ...\left(\text{iii}\right)$

Now, equation of line $AB$

$\overrightarrow{r}=\overrightarrow{a}+k_1\left(\overrightarrow{a}-\overrightarrow{b}\right)\Rightarrow \overrightarrow{r}-\overrightarrow{a}=k_1\left(\overrightarrow{a}-\overrightarrow{b}\right) ...\left(\text{iv}\right)$

$\overrightarrow{r}=\overrightarrow{b}+k_2\left(\overrightarrow{a}-\overrightarrow{b}\right)\Rightarrow \overrightarrow{r}-\overrightarrow{b}=k_2\left(\overrightarrow{a}-\overrightarrow{b}\right) ...\left(\text{v}\right)$

Now, equation of line $CD$

$\overrightarrow{r}=\overrightarrow{c}+k_3\left(\overrightarrow{c}-\overrightarrow{d}\right)\Rightarrow \overrightarrow{r}-\overrightarrow{c}=k_3\left(\overrightarrow{c}-\overrightarrow{d}\right) ...\left(\text{vi}\right)$

$\overrightarrow{r}=\overrightarrow{d}+k_4\left(\overrightarrow{c}-\overrightarrow{d}\right)\Rightarrow \overrightarrow{r}-\overrightarrow{d}=k_4\left(\overrightarrow{c}-\overrightarrow{d}\right) ...\left(\text{vii}\right)$

Putting these values in $\left(\text{iii}\right)$, we get

$$$k_1{k}_2{\left(\overrightarrow{a}-\overrightarrow{b}\right)}^2=k_3{k}_4{\left(\overrightarrow{c}-\overrightarrow{d}\right)}^2 ...\left(\text{viii}\right)$

Now equating $\left(\text{iv}\right) \& \left(\text{vi}\right)$, we get

$\overrightarrow{a}+k_1\left(\overrightarrow{a}-\overrightarrow{b}\right)=\overrightarrow{c}+k_3\left(\overrightarrow{c}-\overrightarrow{d}\right)$

$\Rightarrow \left(1+k_1\right)\overrightarrow{a}-k_1\overrightarrow{b}-\left(1+k_3\right)\overrightarrow{c}+k_3\overrightarrow{d}=0$

Comparing this with equation $\left(\text{i}\right)$, we get

$x=1+k_1, y=-k_1, z=-1-k_3, t=k_3 ....\left(A\right)$

These values are also satisfying equation $\left(\text{ii}\right)$.

Similarly, equating $\left(\text{v}\right) \& \left(\text{vii}\right)$, we get

$\overrightarrow{b}+k_2\left(\overrightarrow{a}-\overrightarrow{b}\right)=\overrightarrow{d}+k_4\left(\overrightarrow{c}-\overrightarrow{d}\right)$

$\Rightarrow k_2\overrightarrow{a}+\left(1-k_2\right)\overrightarrow{b}-k_4\overrightarrow{c}+\left(-1+k_4\right)\overrightarrow{d}=0$

Comparing this with equation $\left(\text{i}\right)$, we get

$x=k_2, y=1-k_2, z=-k_4, t=-1+k_4 ....\left(B\right)$

These values are also satisfying equation $\left(\text{ii}\right)$.

Combining $\left(A\right) \& \left(B\right)$, we get

$k_1=y, k_2=x, k_3=t, k_4=-z$

Putting these values in $\left(\text{viii}\right)$, we get

$-xy{\left(\overrightarrow{a}-\overrightarrow{b}\right)}^2=-zt{\left(\overrightarrow{c}-\overrightarrow{d}\right)}^2$

or 

$\left|xy\right|{\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2=\left|zt\right|{\left|\overrightarrow{c}-\overrightarrow{d}\right|}^2$