Question

Let ABCD be a square. Locate points P, Q, R, S on the sides AB,BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that PQRS is a square.

Answer 1

Let the square ABCD be a rectangle with all equal sides.

Now,
It is given that the points P , Q , R , S are the mid points of AB , BC , CD , DA respectively.


According to the statement :

• AP = BP = AB / 2
• AS = SD = AD / 2
• RD = RC = DC / 2
• BQ = QC = BC / 2



As the given quadrilateral is a rectangle, opposite sides should be equal.
Hence,
• AB = DC

From above,
= > AB / 2 = AP = BP = DC / 2 = RD = RC

= > AP = BP = RD = RC ---: ( 1 )



Also,
• AD = BC

From above,
= > AD / 2 = AS = SD = BC / 2 = BQ = QC

= > AS = SD = BQ = QC ---: ( 2 )



Now, by using Pythagoras Theorem :

In ∆ASP,
• SP² = AP² + AS²

In ∆BPQ,
• QP² = PB² + BQ²

From ( 1 ) & ( 2 ),
= > QP² = ( AP )² + ( AS )²
= > QP² = AP² + AS²
= > QP² = SP²
= > QP = SP


In ∆RDS,
• SR² = SD² + RD²

From ( 1 ) & ( 2 ),
= > SR² = AS² + AP²
= > SR² = SP²
= > SR = SP


In ∆RCQ,
• QR² = QC² + RC²

From ( 1 ) & ( 2 ),
= > QR² = AS² + AP²
= > QR² = SP²
= > QR = SP


Hence,
SR = SP = QR = QP

therefore,
All sides of the formed quadrilateral are equal.



Now, joining the P with D and S with Q,

As AS = BQ = SD = QC, AB || SQ || DC
And, AP = PB = RD = RC, AD || PR || BC


Hence,
Let the intersection of RP & QS be O,

As PB || QO, BQ || OP,
angle PBQ = 90° = angle BPO = angle OPQ,

And,
• angle POQ = angle SOR { vertically opposite angles }

Similarly,
• angle POS = angle QOR = 90°




Hence,
• PQ = PS = SR = RQ

• angle POQ = angle SOR = angle QOR = angle POS = 90°


Hence, the formula quadrilateral is a Rhombus.




Now, when it happens in a square, so :
Where all sides are equal and all the angles are at 90°.


As /_ PBQ = 90° and PB = BQ,
So, /_ BPQ = /_ BQP = 45°

And, doing the same one by one in all triangles.

So,
/_ PSR = /_ SRQ = /_ RQP = /_ QPS = 90°

And, all sides are equal.

Therefore, the formed rhombus is a square.

Proved.

Answer 2

1 ) We know 
AB  =  BC  =  CD  =  DA                     ( As ABCD is a square )
So,

AB  -  AP  =  PB            ---------------- ( 1 )
And
BC  -  BQ  =  QC  

AB  -  AP   =  QC         ---------------- ( 2 )                          ( As we know  AB  =  BC  , and AP  =  BQ is given )

So,
From equation 1 and equation 2 , we get 

PB   =  QC                         ( Hence proved )

2 ) In ∆ PBQ , we apply Pythagoras theorem , As:

PQ2  = PB2  +  BQ2    

PQ  = PB2  + BQ2−−−−−−−−−−√                 -------------------- ( 1 )

And In ∆ RCQ ​ we apply Pythagoras theorem , As:

RQ2 = QC2 +  RC2   

RQ2 = PB2 +  BQ2                                       ( As we know  QC  =  PB  , and RC  =  BQ is given )​

RQ = PB2  + BQ2−−−−−−−−−−√                 -------------------- ( 2 )

So, from equation 1 and 2 we get 

PQ  =   RQ                               ( Hence proved )

3 )  WE know that  
∆ PQR is a isosceles triangle , AS PQ  =  RQ 

SO ∠ RPQ   = ∠ PRQ                             ( Angle opposite of equal sides  )
And
∠ PQR   = 90°                                         ( Given )
And we know

 ∠ RPQ  +  ∠ PRQ  + ​∠ PQR  = 180° 

∠ RPQ  +  ∠ PRQ + ​∠ PQR  = 180° 

2​ ∠ PRQ +  90°   = 180° 

2​ ∠ PRQ  =   90°

∠ PRQ  =   45°                                ( Ans )