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Question (1):
Show that the quadrilateral, formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.
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Answer:
P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
PQRS is a rhombus.
Consider D BAC
( In a triangle the segment joining the mid-points of
two sides are parallel and equal to third side)
Consider D ADC,
From (i) and (ii),
PQRS is a parallelogram ---(iii)
AD = BC ( Opposite sides of a rectange)
i.e. AS = BQ
Consider D APS and D BPQ ,
AP = BP ( P is the mid-point of AB)
AS = BQ
\ PS = PQ ---(iv)
From (iii) and (iv),
PQRS is a parallelogram in which PS = PQ .
 PQRS is a rhombus .
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Question (2):
ABCD is a rhombus and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Prove that PQRS is a rectangle.
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Answer:
ABCD is a rhombus.
P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
PQRS is a rectangle.
Draw the diagonals AC and BD.
Let them intersect at O. Join SP, PQ, QR and RS.
Let SP intersect AC at L and let PQ intersect BD at N.
\ PQRS is a parallelogram.
(One pair of opposite sides are parallel and equal) <
Now consider quadrilateral ONPL
 (diagonals of a rhombus are perpendiculars)
And ON||LP
PN||LO (SP||BD)
(by mid-point theorem)
\ ONLP is a parallelogram in which
one angle = 90o
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Question (3):
The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral, formed by joining the mid-points of its sides is a rectangle.
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Answer:
 .
PQRS is a rectangle.
SR||PQ and SR=PQ (since both SR and PQ are
parallel to AC and equal
(Mid-point theorem)
PQRS is a parallelogram.
Now consider the quadrilateral PLOK,
 .
(One angle of a parallelogram is 90o)
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Question (4):
Show that the quadrilateral formed by joining the mid-points of the sides of a square is also a square.
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Answer:
ABCD is a square. P, Q, R and S are mid-points of AB,
BC, CD and DA respectively.
PQRS is also a square.
Join the diagonals AC and DB. Let SP intersect AC at K and PQ intersect BD at L.
SR||PQ and SR = PQ ( since both SR and PQ are parallel to AC and equal
\PQRS is a parallelogram. (Mid-point theorem)
Now consider the quadrilateral PLOK,
 .
In triangles, APS and PBQ
From (ii) and (iii),
PQRS is a square. ( a rectangle with all four
sides congruent)
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Question (5):
In figure, D, E and F are respectively, the mid-points of sides BC, CA and AB of an equilater at triangle ABC. Prove that D DEF is also an equilateral triangle.
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Answer:
D, E and F are mid-points of BC, CA and AB respectively.
D ABC is an equilateral triangle.
D DEF is also an equilateral triangle.
(segment joining the mid-points of two sides of a triangle is half the third side))<
Similarly,
i.e., ED = EF = DF (from (i), (ii) and (iii))<
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Question (6):
Prove that the figure formed by joining the mid-points of pairs of consecutive sides of a quadrilateral is a parallelogram.
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Answer:
ABCD is a quadrilateral. E, F, G and H are the mid-points of
AB, BC, CD and DA respectively.
EFGH is a parallelogram.
Join A to C and D to B.
Consider D ACD,
mid-points of two sides of a triangle is parallel to the third side and half of it)
Consider D ACB,
mid-points of two sides of a triangle is parallel to the third side and half of it)
From (i) and (ii),
HG=EF and HG||EF
\ EFGH is a parallelogram.
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Question (7):
D ABC is a right angled triangle at B, and P is the mid-point of
 parallel to BC, meeting AB in Q).
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Answer:
D ABC is right angled triangle at B. P is the mid-point of AC.
Through P, draw a line parallel to BC, meeting AB at Q.
In D APQ and D BPQ
PQ = PQ (common side)
PA = PB (corresponding parts of
congruent triangles)
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Question (8):
Prove that the four triangles formed by joining in pairs the mid-points of three sides of a triangle, are congruent to each other.
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Answer:
D, E and F are the mid-points of the sides of D ABC.
 of two sides of a triangle is half the third side)
Similarly DE = BF
In triangles, BDF and DEF
BD = EF (proved above)
BF = DE (proved above)
DF = DF (common side)
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Question (9):
In the figure, D ABC is isosceles with AB=AC, D, E and F are respectively the mid-points of sides BC, AB and AC. Show that AD is perpendicular to the line segment EF and is bisected by it.
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Answer:
AB = AC. D, E and F are the mid-points of BC, BA and AC respectively.
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Question (10):
In figure, E is the mid-point of AD of a trapezium ABCD with AB||DC. A line through E parallel to AB, meets BC in F. Show that F is the mid-point of BC.
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Answer:
E is the mid-point of AD of trapezium ABCD.
AB||DC, EF||AB
F is the mid-point of BC.
AB||DC (given)
EF||AB (given)
i.e., AB, EF and DC are three line segments parallel to each other.
Intercepts AE and ED made by the three parallel segments on AD are equal.
three parallel segment must be equal.
i.e., F is the mid-point of BC.
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Question (11):
Points A, B are on the same side of line l. AD ^ l and BE ^ l in D and E respectively. C is the mid-point of AB. Prove that CD=CE.
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Answer:
D ^ l and BE ^ l.
C is the mid-point of AB.
CD=CE.
Draw CM ^ l.
AD, CM and BE are (  AD, CM and BE are Parallel to each other perpendicular to l)
AC=CB (  C is the mid-point of AB)
 Intercepts made by parallel lines AD,CM and BE on the other transversal l are also equal.
i.e., DM=ME
Consider the triangles, CMD and CME
CM=CM ( common side)
 (  each 900)
DM=ME ( proved above)
 (SAS congruency condition)
 CD=CE (corresponding parts of corresponding triangles)
Hence CD=DE
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Question (12):
E, F are respectively the mid-points of non-parallel sides AD, BC of a trapezium ABCD. Prove that,
a) EF||AB
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Answer:
E and F are mid-points of AD and BC.
a) EF||AB
a)
E and F are mid-points of AD and BC. (given)
DE=AE and CF=BF ( ABCD is a trapezium)
AB||CD
Equal intercepts are made on the transversal AD and BC.
 EF||AB
Hence (a) is proved.
b)
In D ADB, E is the mid-point of AD . (given)
EP||AB ( proved)
 ----(i)
In D BCD, F is the mid-point of BC . (given)
PF||DC
 ----(ii)
By adding (i) and (ii), we get
Hence (b) is proved.
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Question (13):
In D ABC, AD is the median through A and E is the mid-point
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Answer:
AD is the median through A and E is the mid-point of AD.
Through D, draw DK||BF.
Consider D ADK,
 ( given)
 ( by construction)
 ( line drawn through the mid- point of one side of a triangle, parallel to another side,
bisects the third side.)
Consider D BCF,
 (  AD is the median)
DK||BF ( by construction)
 ( line drawn through the mid-point of one side of a triangle parallel to another side, bisects the third side.)
From (i) and (ii),
 (using (iii))
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Question (14):
M and N divide AB of D ABC into three equal parts. Line segment MP and NQ are both parallel to BC and meet AC in P and Q respectively. Prove that P, Q divide AC into three equal parts.
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Answer:
In DABC,
AM=MN=NB
MP||BC and NQ||BC
AP=PQ=QC
Draw DE || BC through A.
AM=MN=NB (given)
and MP||BC; NQ||BC; DE||BC ( given)
i.e., AM, MN and NB are equal intercepts made on transversal AB .
AC is a transversal, intercepts made on AC are AP, PQ and QC.
 AP=PQ=QC
Hence the proof.
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Question (15):
In the adjoining figure, AD||BE||CF and B is the mid-point of AC. CD intersects BE at X. Show that
i) DE = EF
iv) AD + CF = 2BE
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Answer:
AD||BE||CF
AB=BC
CD intersects BE at X.
i) DE=EF
iv) AD+CF=2BE
AD||BE||CF ( given)
DF is a transversal. ( given)
 DF makes equal intercepts.
Since AC makes equal intercepts. ( theorem)
 DE=EF ---(i) ( proved )
B is the mid-point of AC.
X is the mid-point of CD. ( BX||AD)
E is the mid-point of DF . ( proved)
X is the mid-point of BC. ( proved)
BE=BX+EX
 2BE=AD+CF ---(iv) ( proved)
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