Hello, darksoulzero!
$\displaystyle \text{Quadrilateral }ABCD\text{ has vertices at: }A(13, 9),\;B(14, 2),\;C(7, 1),\;D(5, 5).$
$\displaystyle \text{Show that }ABCD\text{ is cyclic.}$
Someone is misinformed.
The diagonals need not be diameters.
Code:
A * * *
o *
*  *
D o+o B

*  *
*  ♥ *
*  O *

*  *
*  *
o *
C * * *
Here is my very primitive solution . . .
We hope to find a point $\displaystyle O(x,y)$ equidistant from $\displaystyle A,B,C,D.$
We have this system of equations:
. . $\displaystyle \begin{array}{ccccccccc}
AO^2 &=& (x13)^2 + (y9)^2 &=& r^2 & [1] \\
BO^2 &=& (x14)^2 + (y2)^2 &=& r^2 & [2] \\
CO^2 &=& (x7)^2 + (y1)^2 &=& r^2 & [3] \\
DO^2 &=& (x5)^2 + (y5)^2 &=& r^2 & [4] \end{array}$
Subtract [1]  [2]: .$\displaystyle (x13)^2  (x14)^2 + (y9)^2  (y2)^2 \:=\:0 $
. . $\displaystyle (1)(2x27) + (\text{}7)(2y11) \:=\:0 \quad\Rightarrow\quad x  7y \:=\:\text{}25\;\;[5]$
Subtract [3]  [4]: .$\displaystyle (x7)^2  (x5)^2 + (y1)^2  (y5)^2 \:=\:0$
. . $\displaystyle (2)(2x12) + (4)(2y6) \:=\:0 \quad\Rightarrow\quad x  2y \:=\:0\;\;[6]$
Solve [5] and [6]: . $\displaystyle x = 10,\;y = 5$
Therefore, $\displaystyle A,B,C,D$ lie on a circle with center $\displaystyle O(10,5)$ and radius 5.