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QUADRILATERAL (2)
Mid-point theorem
The line segment joining the midpoint of any two sides of a triangle is parallel to third side and is half
of it.
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1.
Prove that the figure formed by joinging the midpoints of the consecutive sides of a quadrilateral is a
parallelogram
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3.The diagonal of a quadrilateral ABCD are perpendicular to each other.
Show that the quadrilateral formed by joining the midpoints of its
consecutive sides is a rectangle.
4.ABCD is a rhombus, and P,Q,R, and S are midpoints of the sides AB,
BC,CD and DA respectively. Show that the quadrilateral PQRS is a
rectangle.
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5.Show that the quadrilateral formed by joining the midpoints of the sides of a square is also a
square.
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6.Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect
each other.
7.Show that the quadrilateral formed by joining the midpoints of the consecutive sides of a square
is a square.
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Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of
the sides AB, BC, CD and DA respectively.
However, the diagonals of a square are equal.
∴ AC = BD … (5)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
We know that, diagonals of a square are perpendicular bisector of each other.
∴∠AOD = ∠AOB = ∠COD = ∠BOC = 90°
Now, in quadrilateral EHOS, we have
SE || OH, therefore, ∠AOD + ∠AES = 180° [corresponding angles]
⇒ ∠AES = 180° - 90° = 90°
Again, ∠AES + ∠SEO = 180° [linear pair]
⇒ ∠SEO = 180° - 90° = 90°
Similarly,
SH || EO, therefore, ∠AOD + ∠DHS = 180° [corresponding angles]
⇒ ∠DHS = 180° - 90° = 90°
Again, ∠DHS + ∠SHO = 180° [linear pair]
⇒ ∠SHO = 180° - 90° = 90°
Again, in quadrilateral EHOS, we have
∠SEO = ∠SHO = ∠EOH = 90°
Therefore, by angle sum property of quadrilateral in EHOS, we get
∠SEO + ∠SHO + ∠EOH + ∠ESH = 360°
90° + 90° + 90° + ∠ESH = 360°
⇒ ∠ESH = 90°
In the same manner, in quadrilateral EFOP, FGOQ, GHOR, we get
∠HRG = ∠FQG = ∠EPF = 90°
Therefore, in quadrilateral PQRS, we have
PQ = QR = SR = PS and ∠ESH = ∠HRG = ∠FQG = ∠EPF = 90°
Hence, PQRS is a square.
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8.Prove that the quadrilateral formed by joining the midpoints of the adjacent sides of
rectangle is a rhombus
ABCD is a rhombus. P, Q , R and S are the mid-points of the sides AB, BC, CD and DA
respectively.
Join AC.
In ΔABC, P and Q are the mid points of AB and BC respectively
∴ PQ || AC and PQ =
AC ... (1) (Mid point theorem).
Similarly,
RS || AC and RS =
AC ... (2) (Mid point theorem).
From (1) and (2), we get
PQ || RS and PQ = RS.
Thus, PQRS is a parallelogram (A quadrilateral is a parallelogram, if one pair of opposite sides is
parallel and equal)
AB = BC (Given)
⇒ PB = BQ (P and Q are mid points of AB and BC respectively)
In ΔPBQ,
PB = BQ
∴ ∠BQP = ∠BPQ ... (3) (Equal sides have equal angles opposite to them)
In ΔAPS and ΔCQR,
AP = CQ
AS = CR
PS = RQ (Opposite sides of parallelogram are equal)
∴ ΔAPS
ΔCQR (SSS congruence criterion)
⇒ ∠APS = ∠CQR ... (4) (CPCT)
Now,
∠BPQ + ∠SPQ + ∠APS = 1800
∠BQP + ∠PQR + ∠CQR = 1800
∴∠BPQ+∠SPQ+∠APS=∠BQP+∠PQR+∠CQR
⇒∠SPQ=∠PQR...(5)(from(3)and(4))
PS||QR and PQ is the transversal,
∴∠SPQ+∠PQR=1800(Sumofadjacentinteriorananglesis1800)
⇒∠SPQ+∠SPQ=1800(from(5))
⇒2∠SPQ=1800
⇒∠SPQ=900
Thus, PQRS is a parallelogram such that ∠SPQ=900.
Hence, PQRS is a rectangle
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