1. ## Show that Quadrilateral ABCD is cyclic.

Quadrilateral ABCD has vertices at A(13, 9), B(14, 2), C(7, 1), and D(5, 5). Show that ABCD is cyclic.

I didn't know how to do it without graph paper so I drew the quadrilateral on a graph and said that the diagonals represent the diameter of a circle with the intersection of the diagonals representing the centre of the circle. Half a diagonal also represents the radius of the circle.

I think that this might be correct but I am unsure.

2. Originally Posted by darksoulzero
Quadrilateral ABCD has vertices at A(13, 9), B(14, 2), C(7, 1), and D(5, 5). Show that ABCD is cyclic.

I didn't know how to do it without graph paper so I drew the quadrilateral on a graph and said that the diagonals represent the diameter of a circle with the intersection of the diagonals representing the centre of the circle. Half a diagonal also represents the radius of the circle.

I think that this might be correct but I am unsure.
Derivation / Proof of Ptolemy&#039;s Theorem for Cyclic Quadrilateral

and the converse is also true.

3. Originally Posted by darksoulzero
Quadrilateral ABCD has vertices at A(13, 9), B(14, 2), C(7, 1), and D(5, 5). Show that ABCD is cyclic.

I didn't know how to do it without graph paper so I drew the quadrilateral on a graph and said that the diagonals represent the diameter of a circle with the intersection of the diagonals representing the centre of the circle. Half a diagonal also represents the radius of the circle.

I think that this might be correct but I am unsure.
Be careful here. There is no reason to think that the diagonals of the rectangle are diameters of the circle (although in fact one of them is). It might be best to find the gradients of the lines AB, BC, CD, DA. If the product of two gradients is –1 then those lines are at right angles to each other. And if for example the angle ABC turns out to be a right angle, that would mean that B lies on the circle with diameter AC (because of the theorem that the angle in a semicircle is a right angle, or rather the converse of that theorem).

4. Hello, darksoulzero!

$\text{Quadrilateral }ABCD\text{ has vertices at: }A(13, 9),\;B(14, 2),\;C(7, 1),\;D(5, 5).$
$\text{Show that }ABCD\text{ is cyclic.}$

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 $O(x,y)$ equidistant from $A,B,C,D.$

We have this system of equations:

. . $\begin{array}{ccccccccc}
AO^2 &=& (x-13)^2 + (y-9)^2 &=& r^2 & [1] \\
BO^2 &=& (x-14)^2 + (y-2)^2 &=& r^2 & [2] \\
CO^2 &=& (x-7)^2 + (y-1)^2 &=& r^2 & [3] \\
DO^2 &=& (x-5)^2 + (y-5)^2 &=& r^2 & [4] \end{array}$

Subtract [1] - [2]: . $(x-13)^2 - (x-14)^2 + (y-9)^2 - (y-2)^2 \:=\:0$

. . $(1)(2x-27) + (\text{-}7)(2y-11) \:=\:0 \quad\Rightarrow\quad x - 7y \:=\:\text{-}25\;\;[5]$

Subtract [3] - [4]: . $(x-7)^2 - (x-5)^2 + (y-1)^2 - (y-5)^2 \:=\:0$

. . $(-2)(2x-12) + (4)(2y-6) \:=\:0 \quad\Rightarrow\quad x - 2y \:=\:0\;\;[6]$

Solve [5] and [6]: . $x = 10,\;y = 5$

Therefore, $A,B,C,D$ lie on a circle with center $O(10,5)$ and radius 5.

5. another way is to use this form of circle ${x^2} + {y^2} + Ax + By + C = 0$ Take three points and set them in this equation and you will find A,B,C, solving a simple linear system of equations. If the fourth point satisfies the equation you are done.