Question

ABCD is a square and P, Q,R ,S are midpoints of AB, BC, CD and DA respectively show that PQRS is a square

Answer 1

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Answer 2

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 = $\frac{1}{2}$ AC             [1]

In ΔABC, by mid-point theorem,

 PQ║AC and PQ = $\frac{1}{2}$ AC             [2]

PO║SR and PQ = SR = $\frac{1}{2}$ AC      [ from eq 1 and 2 ]     [3]

Now, in ΔABD, by mid-point theorem,

SP║BD and SP = $\frac{1}{2}$ BD = $\frac{1}{2}$ AC            [4]

In ΔBCD, by mid-point theorem,

RQ║BD and RQ = $\frac{1}{2}$ BD = $\frac{1}{2}$AC            [5]

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