Advertisements
Advertisements
Question
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Advertisements
Solution
Given: In a square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.
To show: PQRS is a square.
Construction: Join AC and BD.
Proof: Since, ABCD is a square.
∴ AB = BC = CD = AD
Also, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.
Then, in ΔADC, SR || AC
And SR = `1/2`AC [By mid-point theorem] ...(i)
In ΔABC, PQ || AC
And PQ = `1/2`AC ...(ii)
From equations (i) and (ii),
SR || PQ and SR = PQ = `1/2`AC ...(iii)
Similarly, SP || BD and BD || RQ
∴ SP || RQ and SP = `1/2`BD
And RQ = `1/2`BD
∴ SP = RQ = `1/2`BD
Since, diagonals of a square bisect each other at right angle.
∴ AC = BD
⇒ SP = RQ = `1/2`AC ...(iv)
From equations (iii) and (iv),
SR = PQ = SP = RQ ...[All side are equal]
Now, in quadrilateral OERF,
OE || FR and OF || ER
∴ ∠EOF = ∠ERF = 90°
Hence, PQRS is a square.
Hence proved.
APPEARS IN
RELATED QUESTIONS
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.
In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 3DF = EF
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]
