Question

ABCD is a convex quadrilateral. Let P, Q, R, S be centers of the squares drawn (externally to ABCD) on AB, BC, CD, DA respectively. Prove that P R is equal and perpendicular to QS.​

Answer 1

Step-by-step explanation:

Given- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To Prove-PQRS is a rectangle

Construction,

AC and BD are joined.

Proof,

In ΔDRS and ΔBPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

∠SDR = ∠QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.

RS = PQ by CPCT --- (i)

In ΔQCR and ΔSAP,

RC = PA (Halves of the opposite sides of the rhombus)

∠RCQ = ∠PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.

RQ = SP by CPCT --- (ii)

Now,

In ΔCDB,

R and Q are the mid points of CD and BC respectively.

⇒ QR || BD

also,

P and S are the mid points of AD and AB respectively.

⇒ PS || BD

⇒ QR || PS

Thus, PQRS is a parallelogram.

also, ∠PQR = 90°

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

∠Q = 90°

Thus, PQRS is a rectangle.

Answer 2

Given:

ABCD is a quadrilateral, P, Q, R, S center of a square drawn externally to ABCD.

To Find:

PR is equal to and perpendicular to QS.

Solution:

we need to construct, AC and BD are joined.

Now,

In ΔDRS and ΔBPQ,

DS is equal to BQ as they are the halves of the rhombus.

∠SDR = ∠QBP   [opposite angles of the rhombus]

DR = BP [halves of the opposite side of rhombus]

Thus,

   ΔDRS≅ ΔBPQ  [SAS congruence rule]

     RS = PQ [C.P.C.T]..(i)

      In ΔQCR and ΔSAP,

       RC = PA ( halves of the opposite sides of the rhombus)

      ∠RCQ = ∠PAS(opposite angles of the rhombus)

       CQ = AS (halves of the opposite sides of the rhombus)

thus,

       ΔQCR ≅ ΔPAS   [ SAS congruence rule]

         RQ = SP [C.P.C.T]..(ii)

     Now, ΔCDB,

             R and Q are the mid-points of CD and BC respectively.

         ⇒ QR ║ BD

   also,

P and S are the mid-points of AD and AB respectively.

⇒PS║BD

⇒QR║PS

Thus, PQRS is a parallelogram.

∠PQR = 90°

Now, in PQRS,

RS = PQ and RQ = SP (from i and ii)

∠Q = 90°

  Hence, PQRS is a rectangle.