Question

Heya!


`\underline{ \underline{ \text{question}}}:` In the given figure, PQRS is a square. A , B , C and D are the points on the sides PQ , QR , RS and SP respectively. If AQ = BR = CS = DP , prove that ABCD is also a square.

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Answer 1

the answer is in last picture

Answer 2

See Below.

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:

`m\angle P=m\angle Q=m\angle R=m\angle S=90^\circ`

Likewise:

`PQ=QR=RS=SP`

PQ is the sum of the segments PA and AQ:

`PQ=PA+AQ`

Likewise, QR is the sum of the segments QB and BR:

`QR=QB+BR`

Since AQ = BR and PQ = QR:

`PA+AQ=QB+AQ`

Therefore:

`PA=QB`

Likewise, RS is the sum of the segments RC and CS:

`RS=RC+CS`

Since RS = PQ and CS = AQ:

`RC+CS=PA+CS`

Thus:

`RC=PA=QB`

Repeating this procedure for the remaining side, we acquire that:

`PA=QB=RC=SD`

And since each of the angles ∠P, ∠Q, ∠R, and ∠S is 90°, by the SAS Theorem, we acquire:

`\Delta AQB\cong\Delta BRC\cong \Delta CSD\cong \Delta DPA`

Then by CPCTC, we acquire:

`AB=BC=CD=DA`

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:

`m\angle QAB+m\angle QBA=90^\circ`

By CPCTC, ∠PAD≅∠QBA. Thus:

`m\angle QAB+m\angle PAD=90^\circ`

∠DAB forms a linear pair. Therefore:

`m\angle PAD+m\angle QAB+m\angle DAB=180`

By substitution and simplification:

`m\angle DAB=90^\circ`

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.