Question

in the rectangle abcd shown, x and y are mid - points of the given sides and dp = aq. 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: Identify given congruent parts

We know that in rectangle \(ABCD\), \(\overline{DC}\cong\overline{BA}\), \(\overline{PD}\cong\overline{AQ}\), so \(\overline{CP}\cong\overline{BQ}\) (by subtraction property of congruence). Also, \(\overline{CY}=\overline{BY}\) and \(\angle C=\angle B = 90^{\circ}\).

Step2: Recall congruence - postulates

The Side - Angle - Side (SAS) postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent. In \(\triangle CYP\) and \(\triangle BYQ\), we have \(\overline{CP}\cong\overline{BQ}\), \(\angle C=\angle B\) and \(\overline{CY}=\overline{BY}\).

Answer:

A. \(\triangle CYP\cong\triangle BYQ\) by SAS