Since we are given [math]\overline{PQ}\parallel\overline{BD}[/math], we know, by applying the Side-Splitter Theorem to [math]\triangle ABD[/math], that [math]\overline{PQ}[/math] divides [math]\overline{AB}[/math] and [math]\overline{AD}[/math] proportionally: [math]\frac{AP}{PB}=\frac{AQ}{QD}[/math].[br][br]In [math]\triangle ADC[/math], since [math]\overline{QR}\parallel\overline{AC}[/math], by the Side-Splitter Theorem, [math]\overline{QR}[/math] divides [math]\overline{AC}[/math] and [math]\overline{CD}[/math] proportionally: [math]\frac{AQ}{QD}=\frac{CR}{RD}[/math].[br][br]In [math]\triangle ABC[/math], since [math]\overline{PS}\parallel\overline{AC}[/math], by the Side-Splitter Theorem, [math]\overline{PS}[/math] divides [math]\overline{AB}[/math] and [math]\overline{BC}[/math] proportionally: [math]\frac{AP}{PB}=\frac{CS}{SB}[/math].[br][br]By the transitive property of equality, we therefore have that [math]\frac{CS}{SB}=\frac{CR}{RD}[/math]. In other words, in [math]\triangle BCD[/math], we have that [math]\overline{RS}[/math] divides [math]\overline{BC}[/math] and [math]\overline{CD}[/math] proportionally. In a triangle, a line that divides two sides proportionally is parallel to the third side, so [math]\overline{RS}\parallel\overline{BD}[/math][br][br]Finally, since two lines parallel to the same line are parallel to each other, as [math]\overline{PQ}\parallel\overline{BD}[/math] and [math]\overline{RS}\parallel\overline{BD}[/math], we reach the conclusion [math]\overline{PQ}\parallel\overline{RS}[/math].[br][br]Note: Since two lines parallel to the same line are parallel to each other, as [math]\overline{PS}\parallel\overline{AC}[/math] and [math]\overline{QR}\parallel\overline{AC}[/math], we have [math]\overline{PS}\parallel\overline{QR}[/math], which means [math]PQRS[/math] will always be a parallelogram, since both pairs of opposite sides are parallel.

[math]ABCD[/math] can be any quadrilateral![br][br][math]PQRS[/math] will always be a parallelogram, as proven above.[br][br]If [math]P[/math] is close to [math]A[/math], then [math]PQ, but if [math]P[/math] is close to [math]B[/math], we get [math]PQ>QR[/math]. As we move [math]P[/math] closer to [math]B[/math], the length of [math]\overline{PQ}[/math] increases, which means that, somewhere on [math]\overline{AB}[/math] there is a point [math]P[/math] where [math]\overline{PQ} \cong \overline{QR}[/math].[br][br]Since [math]PQRS[/math] is a parallelogram with two congruent adjacent sides, it is a rhombus.

As proven, [math]PQRS[/math] will always be a parallelogram.[br]As long as [math]\overline{AC}\perp\overline{BD}[/math], then [math]\overline{PQ}\perp\overline{BD}[/math], because if two lines are parallel, a line perpendicular to one will be perpendicular to the other.[br][br]If two lines are a parallel, a line perpendicular to one is perpendicular to the other, so since [math]\overline{BD}\parallel\overline{QR}[/math] and [math]\overline{PQ}\perp\overline{BD}[/math], which means we will have [math]\overline{PQ}\perp\overline{QR}[/math], making [math]\angle PQR[/math] a right angle. [br][br]A parallelogram with a right angle is a rectangle, so [math]PQRS[/math] is a rectangle.