Answer:
See Below.
Step-by-step explanation:
We are given that PQRS is a square. A, B, C, and D are the points of the sides PQ, QR, RS, and SP, respectively.
And we are given that AQ = BR = CS = DP.
This is shown in the figure below.
And we want to prove that ABCD is a square.
Since PQRS is a square, it follows that:
[tex]m\angle P=m\angle Q=m\angle R=m\angle S=90^\circ[/tex]
Likewise:
[tex]PQ=QR=RS=SP[/tex]
PQ is the sum of the segments PA and AQ:
[tex]PQ=PA+AQ[/tex]
Likewise, QR is the sum of the segments QB and BR:
[tex]QR=QB+BR[/tex]
Since AQ = BR and PQ = QR:
[tex]PA+AQ=QB+AQ[/tex]
Therefore:
[tex]PA=QB[/tex]
Likewise, RS is the sum of the segments RC and CS:
[tex]RS=RC+CS[/tex]
Since RS = PQ and CS = AQ:
[tex]RC+CS=PA+CS[/tex]
Thus:
[tex]RC=PA=QB[/tex]
Repeating this procedure for the remaining side, we acquire that:
[tex]PA=QB=RC=SD[/tex]
And since each of the angles ∠P, ∠Q, ∠R, and ∠S is 90°, by the SAS Theorem, we acquire:
[tex]\Delta AQB\cong\Delta BRC\cong \Delta CSD\cong \Delta DPA[/tex]
Then by CPCTC, we acquire:
[tex]AB=BC=CD=DA[/tex]
However, this means that ABCD could be either a rhombus or a square, so we need to prove that the angles are right angles.
The interior angles of a triangle always sum to 180°. Since they are right triangles, the two other angles must equal 90°. Thus, for ΔAQB:
[tex]m\angle QAB+m\angle QBA=90^\circ[/tex]
By CPCTC, ∠PAD≅∠QBA. Thus:
[tex]m\angle QAB+m\angle PAD=90^\circ[/tex]
∠DAB forms a linear pair. Therefore:
[tex]m\angle PAD+m\angle QAB+m\angle DAB=180[/tex]
By substitution and simplification:
[tex]m\angle DAB=90^\circ[/tex]
So, ∠DAB is a right angle.
By repeating this procedure, we can establish that ∠ABC, ∠BCD, and ∠CDA are all right angles. This is not necessary, however. Since we concluded that AB = BC = CD = DA, and that we have one right angle, then ABCD must be a square.
