Question

if ABCD is a square and P,Q,R,S are the mid points of AB, BC, CD and DA respectively prove that PQRS is also a square ( Hint : Prove PS = PQ and angle PSR = 90°​

Answer 1

Step-by-step explanation:

PQRS is a square.

Proved below.

Step-by-step explanation:

Given:

P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD and AC is perpendicular to BD. [diagonals of a square are equal]

In ΔADC, by mid-point theorem,

SR || AC and SR = \frac{1}{2}

2

1

AC [1]

In ΔABC, by mid-point theorem,

PQ║AC and PQ = \frac{1}{2}

2

1

AC [2]

PO║SR and PQ = SR = \frac{1}{2}

2

1

AC [ from eq 1 and 2 ] [3]

Now, in ΔABD, by mid-point theorem,

SP║BD and SP = \frac{1}{2}

2

1

BD = \frac{1}{2}

2

1

AC [4]

In ΔBCD, by mid-point theorem,

RQ║BD and RQ = \frac{1}{2}

2

1

BD = \frac{1}{2}

2

1

AC [5]

SP = RQ = \frac{1}{2}

2

1

AC [ from eq 4, 5] [6]

PQ = SR = SP = RQ [ from eq 3, 5]

Thus, all four sides are equal.

Now, in quadrilateral EOFR,

OE║FR, OF║ER

∠ EOF = ∠ ERF = 90° (Opposite angles of parallelogram)

∠ QRS = 90°

Hence, PQRS is a square

Answer 2

Answer:

Step-by-step explanation:

P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD and AC is perpendicular to BD.       [diagonals of a square are equal]

In ΔADC, by mid-point theorem,  

SR || AC and SR =  AC             [1]

In ΔABC, by mid-point theorem,

PQ║AC and PQ =  AC             [2]

PO║SR and PQ = SR =  AC      [ from eq 1 and 2 ]     [3]

Now, in ΔABD, by mid-point theorem,

SP║BD and SP =  BD =  AC            [4]

In ΔBCD, by mid-point theorem,

RQ║BD and RQ =  BD = AC            [5]

SP = RQ =  AC                           [ from eq 4, 5]                [6]

PQ = SR = SP = RQ                        [ from eq 3, 5]

Thus, all four sides are equal.

Now, in quadrilateral EOFR,  

OE║FR, OF║ER

∠ EOF = ∠ ERF = 90°           (Opposite angles of parallelogram)

∠ QRS = 90°  

Hence, PQRS is a square.