A square $P_1P_2P_3P_4$ has points $X$ on side $P_2P_3$ and $Y$ on side $P_3P_4$ chosen such that angle $XP_1Y$ equals forty-five degrees. The lines $P_1X$ and $P_1Y$ intersect the circumcircle of the square at points $R$ and $S$, respectively.
How can one show that the lines $XY$ and $RS$ run parallel?
A geometric-algebraic solution using trigonometric addition properties is fairly straightforward and gives a solution. Yet here I´m looking for a more elegant elementary geometric solution. The intersections with the circumcircle, points $R$ and $S$ can be construed as corners of another square of equal size as the initial one using the Pythagorean theorem.
Here´s where I think a line or two or a smart angle hunt could solve it but got stuck. Any suggestion?
Consider the quadrilateral RSYX and show that it is an isosceles trapezoid. You can show it by proving that SY and RX have the same length and SYX and YXR form the same angle.