IF ABCD is a square, and P, Q, R, S are the mid points of AB, BC, CD and DA respectively. Prove that
PORS is also a square. (Hint: Prove PS = PQ and ZPSR = 90°)
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S
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Answers
Answer:
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Step-by-step explanation:
Given: ABCD is a rhombus. P, Q, R, S are mid points of AB, BC, CD, DA respectively.
Join: AC and BD
In △ABC
P is mid point of AB and Q is mid point of AC.
Thus, by mid point theorem, PQ∥AC and PQ=
2
1
AC
Similarly, In △ACD,
S is mid point of AD and R is mid point of CD
Thus, by mid point theorem, SR∥AC and SR=
2
1
AC
Hence, PQ∥SR and PQ=SR
Similarly, PS=QR and PS∥QR
Thus, the opposite sides of PQRS are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since AC⊥BD thus, PS⊥PQ (Angle between two lines is same as the angle between their corresponding parallel lines)
Now, the opposite sides of PQRS are equal and parallel and the sides meet each other at right angles. Hence, PQRS is a rectangle.
If PQRS had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if ABCD 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.
ANSWERED BY GAUTHMATH