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°
Answers
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:
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.