Question

in the rectangle abcd shown, x and y are mid - points of the given sides and dp = aq.
a) what type of quadrilateral is pyqx? prove your answer.
b) if points p and q are moved at a constant rate and in the same direction along dc and ab, respectively, does this change your answer in part a)? why or why not?

b. △pdx≅△qax by sas
c. △pdx≅△qax by sss
d. △cyp≅△byq by sas

px≅qx by cpctc. dc≅ba as opposite sides of a rectangle. since pd≅qa, then cp≅bq. given also that cy = by and ∠c and ∠b are right angles, which of the following can be concluded?
a. △cyp≅△byq by sas
b. △cyp≅△byq by hl
c. △cyp≅△byq by aas
d. △cyp≅△byq by sss

Explanation:

Step1: Recall rectangle properties

In rectangle \(ABCD\), \(AB = DC\), \(AD=BC\), and \(\angle A=\angle B=\angle C=\angle D = 90^{\circ}\). Given \(X\) is mid - point of \(AD\) and \(Y\) is mid - point of \(BC\), so \(AX=XD\) and \(BY = YC\). Also, \(DP = AQ\).

Step2: Prove congruent triangles for part a

In \(\triangle DXP\) and \(\triangle AXQ\), \(AX = XD\) (mid - point of \(AD\)), \(\angle D=\angle A = 90^{\circ}\), and \(DP = AQ\). By the Side - Angle - Side (SAS) congruence criterion, \(\triangle DXP\cong\triangle AXQ\). So \(PX=QX\). Similarly, in \(\triangle CYP\) and \(\triangle BYQ\), \(CY = BY\), \(\angle C=\angle B = 90^{\circ}\), and since \(DC = AB\) and \(DP = AQ\), then \(CP=BQ\). By SAS congruence criterion, \(\triangle CYP\cong\triangle BYQ\), so \(PY = QY\). Also, \(XY\) is parallel to \(AB\) and \(DC\) (as \(X\) and \(Y\) are mid - points of \(AD\) and \(BC\) respectively). Since \(PX = QX\) and \(PY = QY\), quadrilateral \(PYQX\) is a parallelogram (a quadrilateral with both pairs of opposite sides equal).

Step3: Analyze part b

If \(P\) and \(Q\) are moved at a constant rate and in the same direction along \(DC\) and \(AB\) respectively, the equalities \(AX = XD\), \(BY = YC\), \(\angle A=\angle B=\angle C=\angle D = 90^{\circ}\), and the relationship \(DC = AB\) still hold. Also, the equalities \(DP = AQ\) (even as they change, the difference between the lengths from the endpoints of \(DC\) and \(AB\) remains the same) will ensure that the congruence of \(\triangle DXP\cong\triangle AXQ\) and \(\triangle CYP\cong\triangle BYQ\) still holds. So the quadrilateral \(PYQX\) will still be a parallelogram.

Step4: Answer the multiple - choice question

In \(\triangle CYP\) and \(\triangle BYQ\), we have \(CY = BY\) (mid - point of \(BC\)), \(\angle C=\angle B = 90^{\circ}\), and \(CP=BQ\) (because \(DC = AB\) and \(DP = AQ\)). By the Side - Angle - Side (SAS) congruence criterion, \(\triangle CYP\cong\triangle BYQ\).

Answer:

a) Quadrilateral \(PYQX\) is a parallelogram. Proof: We proved \(\triangle DXP\cong\triangle AXQ\) and \(\triangle CYP\cong\triangle BYQ\) which gives \(PX = QX\) and \(PY = QY\). Also, \(XY\) is parallel to \(AB\) and \(DC\). A quadrilateral with both pairs of opposite sides equal is a parallelogram.
b) No, it does not change the answer. The properties of the rectangle and the equal - length relationships still hold, so the congruence of the relevant triangles still holds, and the quadrilateral \(PYQX\) remains a parallelogram.
For the multiple - choice question: A. \(\triangle CYP\cong\triangle BYQ\) by SAS