parmenides51 wrote: Let $ABCD$ be a quadrilateral that can be inscribed in a circle whose diagonals are perpendicular. Denote by $P$ and $Q$ the feet of the perpendiculars through $D$ and $C$ respectively on the line $AB$, $X$ is the intersection point of the lines $AC$ and $DP$, $Y$ is the intersection point of the lines $BD$ and $CQ$. Show that $XY CD$ is a square. $XYCD$ is a rhombus.

indeed, in the official wording it stated square (cuadrado) while in the official solution, it stated rhombus (rombo) I have just corrected it, thanks for noticing Quote: Let $ABCD$ be a quadrilateral that can be inscribed in a circle whose diagonals are perpendicular. Denote by $P$ and $Q$ the feet of the perpendiculars through $D$ and $C$ respectively on the line $AB$, $X$ is the intersection point of the lines $AC$ and $DP$, $Y$ is the intersection point of the lines $BD$ and $CQ$. Show that $XY CD$ is a square rhombus.