Hello, GreenDay14!

I have an analytic proof . . .

Through a point on the diagonal of a square,

lines $\displaystyle PR, QS$ are drawn parallel to the sides; $\displaystyle P, Q, R, S$ being on the sides.

Prove that these four points are concyclic. Code:

| Q
(0,a)| (b,a) (a,a)
- - - - - - o - -
| | * |
P | | * | R
(0,b) o - - - - - - * - - o (a,b)
| * | |
| o | |
| * C | |
| * | |
| * | |
| * | |
- - - - - - - - - o - - - - -
(0,0) (b,0) (a,0)
S

Let $\displaystyle a$ = side of the square.

Let $\displaystyle (b,b)$ be the point on the diagonal.

Then we have: .$\displaystyle P(0,b),\;Q(b,a),\;R(a,b),\;S(b,0)$

Let $\displaystyle C\left(\tfrac{a}{2},\:\tfrac{a}{2}\right)$ be the center of the square.

Find the distances: .$\displaystyle CP,\:CQ,\:CR,\:CS$

. . $\displaystyle CP^2 \:=\:\left(\tfrac{a}{2}-0\right)^2 + \left(\tfrac{a}{2}-b\right)^2 $

. . $\displaystyle CQ^2 \:=\:\left(\tfrac{a}{2}-b\right)^2 + \left(a - \tfrac{a}{2}\right)^2$

. . $\displaystyle CR^2 \:=\:\left(\tfrac{a}{2}-a\right)^2 + \left(\tfrac{a}{2}-b\right)^2$

. . $\displaystyle CS^2 \:=\:\left(\tfrac{a}{2}-b)\right)^2 + \left(\tfrac{a}{2} - 0\right)^2$

We find that they are *all* equal to: .$\displaystyle \tfrac{1}{2}a^2 - ab + b^2$

Hence, there is a point $\displaystyle C$ equidistant from $\displaystyle P,Q,R,S.$

. . $\displaystyle C$ is the circumcenter of quadrilateral $\displaystyle PQRS.$

Therefore, $\displaystyle P,Q,R,S$ are concyclic.