Chapter 11
MID – POINT THEORAM Exercise 11
Q.1 (a) IN the figure (1) given below, D, E and F are mid-point of the sides BC, CA and AB respectively Of ABC. IF AB =6 cm, BC = 4.8 cm, and CA= 5.6 cm, find the perimeter of (i) trapezium FBCE
(ii) the triangle DEF.
(b) In the figure (2) given the below, D and E are mid-point of the sides AB and AC respectively. If BC= 5.6 cm and B 72 , compute (i) DE (ii) ADE.
(c) In the figure (3) given below, D and E are midpoints of AB, Bc respectively. Calculate AC if AF = 2.6 cm.
Ans. (a) (i) Given : AB=6 cm, BC = 4.8 cm, and CA = 5.6 cm Required: The perimeter of trapezium FBCA.
Sol. F is the mid-point of AB
1 1
6 3
2 2
BF AB cm
intof E is the midpo AC
1 1
5.6 2.8 ... 2
2 2
CE AC cm cm
Now F and E are the mid-point of the AB and CA 1
FE BC abnd FE 2 BC
1
4.8 2.4 .... 3
FE 2 cm cm
Perimeter of the trapezium FBCE
[substituting the value from (1), (2) and (3)]
3 4.8 2.8 2.4 13
BF BC CF EF
cm cm cm cm cm
Hence, perimeter of trapezium DFBCE = 13cm cm Ans.
(ii) D, E and F are the mid-point of the sides BC, CA and AB of ABC respectively. 1
EF BCandEF 2BC
1 4.8 2.4 cm EF 2
Similarly,
1 1
6 3
2 2
1 1
5.6 8.2
2 2
3 2.4 2.8 8.2 .
DE AB cm cm
and FD AC cm cm
PErimeter of DEF DEEF FD
cm cm cm cm Ans
(b) Given: D and E are mid-point of the sides AB and AC respectively. BC =5.6 cm and 72
B
Required: (i) DE (ii) ADE Sol. In ABC
D and E is mid-point of the sides AB and AC respectively.
BY mid-point theorem 1 DE BC and DE 2BC
(i) 1 1 5.6 2.8
2 2
DE BC cm cm (ii) ADE B
[(corresponding angles )]
72
ADE ` [BC DE]
[ B 72 (given) ]
(c) Given: D and E are mid-points of AB, BC respectively and DF BC AF, 2.6 cm. To prove: (i) BEF is a parallelogram
(ii). To calculate the value of AC
Proof: (i) In int
int ...(1)
ABC
D is the mid po of AB and Df BC F is the mid po of AC
Now F and E are mid-point of AC and BC respectively.
... 2 , DF BC
DF BE ... 3
2 ... 4 EF Ab
Now
EF AB From
EF DB
From (3) and (4), DBEF is a parallelogram (iii) F is mid-point of AC
2 2 2.6 5.2 .
AC AF cm cm
Q.2 Prove that the four triangle formed by joining in pair the mid-point of the sides of a triangle are congruent to each other.
Ans. Given: InABC D E, , and F are mid-points of AB, BC and CA respectively. Join DE,EF and FD.
To prove:
. ADF DBE ECF DEF
Proof: In ABC , D and E are mid-point of AB and BC respectively DE AC or FC
Similarly, DF EC
DECF is a parallelogram.
Diagonal FE divides the parallelogram DECF in two congruent triangle DEF and EF.
... 1 DEF FCF
Similarly we can prove that,
... 2 .... 3 DBE DEF
and DEF ADF
From (1),(2) and (3),
ADF DBE ECF DEF
Q.3 If D, E and F are mid-point of sides AB, BC CA respectively of an isosceles triangle ABC<
prove that DEF is also isosceles.
Ans. Given: ABC is an isosceles triangle in which AB=AC d, E and F are mid-point of the sides DC,CA and AB respectively D, E, F are joined
To prove: DEF is an isosceles triangle.
Proof: D and E are the mid-point of BC and AC 1
... 1 DE AB and DE 2AB
Again, D and F are the mid-point of BC and AB respectively.
1 ... 2
DF AC andDF 2AC
AB BC given
DE DF
DEF is an isosceles traingle
Q.4 The diagonals AC and BD o a parallelogram ABCD intersect at O. If P is the mid-point of AD,
Prove that (i) PQ AB (ii) PO 1
2CD
Ans. (i) Given: ABCD is a parallelogram in which diagonals AC and BD intersect each other. AT point O, P is the mid-point of AD. Join OP.
To Prove: (i) (ii) PQ 1 . PQ AB 2CD
Proof: We know that in parallelogram diagonals bisect each other.
BO OD
i.e. O is the mid-point of BD
Now, in ABD,
P and O is the mid-point of AD and BD respectively
1 ...(1)
PO AB and PO 2AB
i.e. PO AB
(ii) Now ABCD is a parallelogram
... 2 AB CD
From (1) and (2), 1
PO2CD
Q.5 In the adjoining figure, ABCD is a quadrilateral in which P,Q, R and S mid-point of AB, BC CD and DA respectively. AC is it =diagonal. Show that
(i) 1
SR AC and SR 2AC (ii) PQSR
(iii) PQRS is a parallelogram.
Sol. Given: In quadrilateral ABCD P, Q, R and S are the mid-point of sides AB, BC, CD and DA respectively AC is the diagonal.
To prove: (i) SR 1
AC and SR2AC (ii) PQ+SR
(iii).PQRS is a Parallelogram
Proof: (i) In ADC S and R are the mid-point of AD and DC 1
2 ....
SR AC and SR AC i
(ii) Similarly in ABC.
P and Q are mid-point of AB and BC 1
...
PQ AC and PQ 2AC ii From (i) and (ii)
PQ = SR and PQ SR (iii) PQSR and PQ SR
log PQRS is a paralle ram
Q.6 Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.
Ans. Given: A square ABCD in Which E,F,G and H are mid-point of AB, BC, CD, and DA respectively join EF, FG and HE.
To Prove: EFGH is a squre Construction: Join AC and BD
Proof: In ACH,G and H are mid-point Of CD and AC respectively.
1
... 1 GH ACandGH 2AC
Now in ABC,E and F are mid-point of AB and Bc respectively.
1
... 2 EF ACandEF 2AC
From (1) and (2),
1
... 3 EF GH and EF GH 2AC
Similarly, we can prove that 1 EF GH and EHGF 2BD
But AC =BD ( Diagonal of square are equal) Dividing both sides by 2,
1 1
... 4 2AC 2BD
From (3) and (4)
EF = GH = EH = GF ….(5)
EFGH is a parallelogram Now, in GOH and GOF
OH = OF (Diagonal of parallelogram bisect each other)
OG = OG (common)
GH = GF [from (5)]
GOH GOF
[By S.S.S. axiom of congruency ]
(c.p.c.t) Now GOH GOF 180
180 2
180 90 2
GOH GOF
Linearpair
or GOH GOH
or GOH
GOH
Diagonals of parallelogram ABCD bisect and perpendicular to each other.
EFGH is a square
Q.7 IN the joining figure, AD and BE are medians of ABC . If DF DE, prove tha
1 .
CF 4 AC
Sol. Given: IN the given figure AD and BE are the medians of ABC DF BE is drawn To Prove: 1
CF 4AC Proof: In BCE
D is the mid-point of BC and DF BE
F is the mid-point of EC 1
2 ...
CF EC i
E is the mid-point of AC
1 ...(ii)
EC 2AC
From (i) and (ii),
1 1 1 1
2 2 2 4
CF EC AC AC
Q.8 (a) In the figure (1) given below, ABCD is a parallelogram. E and F are mid-point of the sides AB and CD respectively. They straight lines AF and BF meet the Straight lines ED and EC in point G and H respectively. Prove that
(i) HEB HCF (ii) GEHF is an parallelogram.
(b) In the diagram (2) given below, ABCD is a parallelogram. E is mid-point of CD and P is a point on AC such that 1 .
PC 4AC EP produced meets BC at F. Prove that (i) F is mid-point of BC (ii) 2Ef = BD
Ans. Given: ABCD is a parallelogram. E and F are mid-point of the side AB and CD repectively.
To Prove:
(i)
(ii) log .
HEB HCF
GEHF is a paralle ram FC BE
... 1
... 2
CEB FCE Alternate angles HEB FCH
ALSO EBF CFB Alternate angles EBH CFM
E and F are mid-points of AB and CD respectively.
1 .... 3
2
1 .... 4
2 BE AB
and CF CD
But ABCD is a parallelogram AB CD
1 1 1
2 2 2
3 4 .... 5
, in HEB and HCF
HEB FCH (1)
2 5 AB CD Dividingbothsidesby
BE CF From and
Now
From
EBH CFH From
BE CF FRom
HEB HCF
(By A.S.A axiom of congruency) (ii) Since E and F are mid-point of AB and CD
AE CF AB CD
Now AE CF .
AE CF and AE CF AECF is a gm
Now, g and H point on the AF and CE respectively.
....(6) GF EH
Similarly we can prove that GFHE is all a gm.
Now point G and H on the line DE and BF respectively .
.... 7 GE EH
From (6) and (7)
GEHF is a parallelogram.
(b) Give: ABCD is a parallelogram in which E is the mid-point of DC and P is a point on AC such the mid-point of DC and P is a point on AC such that 1 .
PC 4AC EP produced meets BC at F.
To Prove: (i) F is the mid-point of BC (ii). 2EF = BD
Construction: Join BD to intersect AC at O.
Proof: Diagonals of gm bisect each other.
1 4
1 1
4 2 2
AO CO But CP AC
Cp CO CP CO
i.e. P is mid-point of CO.
In COD E and P are mid , point of DC & Co.
. .
EP DO i e EF DO
Further, in CBD, E in mid-point of DC and EF BD
Fis the mid-point of BC and 1 EF 2BD . . 2
i e EF BD
Q.9 ABC is an isosceles triangle with AB=AC. D, E and F are mid-point of the sides BC, AB and AC respectively. Prove that of the segment AD is perpendicular to EF and I bisected by it.
Ans. Given: ABC is an isosceles triangle with AB =AC. D, E and F are mid-point of the sides BC, AB and AC respectively.
To Prove: ADEF and AD bisect the EF.
Proof: In ABD and ACD ABD ACD
(ABC is an isosceles triangle)
BD = BC (given D is mid-point of BC)
AB = BC (given)
ABD ACD
180 180
2 180 90
2
... 1 ADB AOC
Also ADB ADB ADB ADB
ADB ADB
AD BC
Now D and E are mid-point of BC and AB (given) DE AF ... 2
Again D and F are mid-point of BC and AC
... 3 EF AD
From (2) and (3) . AEDF is a gm
Diagonals of a parallelogram bisect each other
AD and EF bisect each other From (1) and (3)
ADEF EF BC
Q.10 (a) In a quadrilateral (1) given below, AB DC, E and F are mid-point of Ad and BD respectively. Prove that:
(i) G is mid-point of BC (ii) 1
EG2 ABDC
(b) In the quadrilateral (2) given below, AD DC EG. If E is mid-point of AD prove that:
(i) G is mid-point of BC (ii) 2EG+AB+CD.
(c) IN the quadrilateral (3) given the below, AB DC. E and F are mid-point of non- parallel sides Ad and BC respectively. Calculate:
(i) EF if AB = 6 cm and DC = 4cm (ii) AB if DC = 8cm and EF = 9 cm.
Ans. (a) Given: AB DC E, and F are mid-point of AD and BD respectively
To Prove:
(i) G is mid-point of BC
(ii) 1
.EG 2 ABDC Proof:
In ABD DF, BF
Fismidpointof BD
Also E is the mid-point of AD
1 ... 1
EF AB and EF 2AB
EG CD AB CD given
Now F is mid-point of BD and FG DC
G is mid-point of BC.
1
... 2 FG 2DC
Adding (1) and (2), we get
1 1 1
2 2 2
EFFG AB DC EG ABDC
Hence, the result.
(b) Given: Quadrilateral ABCD in which AB DC EG E. is mid-point of AD.
To Prove: (i) G is mid-point of BC.
(ii). 2EG = AB +CD
Proof: AB DC (given) and EG DC
given
, EG DC In DAB
E is mid-point of AD and EG AB
F is the mid-point of BD and 1 ....(1) EF 2AB
In BCD,
F is mid-point of BD and FG DC
1 ... 2
Fg 2CD
Adding (1) and (2)
1 1
2 2
1 2
EF Fg AB CD
EG AB CD
(c) Given: A quadrilateral in which AB DC E, and F are mid-point of non parallel sides AD and BC respectively.
Required: (i) EF if AB = 6 cm and DC = 4cm (ii). AB if DC = 8 cm and EF = 9 cm
Now, the length of line segment joining the mid-points of two non-parallel sides is half the sum of the length of the parallel sides.
E and F are mid-point of AD and BC respectively.
1 ... 1
EF 2 AB CD
(i). AB = 6 cm and DC = 4 cm Putting these in (1), we get
1 1
6 4 10 5 .
2 2
EF cm Ans
(ii). DC = 8cm and EF = 9cm Putting these in (1), we get
1 1
9 8
2 2
18 8 18 8
10 .
EF AB DC AB
AB AB
AB cm Ans
Q.11 (a) In the quadrilateral (1) given below, AD = BC p, Q, R and S are mid-point of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.
(b) In the figure (2) given below, ABCD is a kite in which BC=CD, AB ,E,F,G are mid- point of CD, BC and AB respectively. Prove that:
(i) EFG 90
(ii) The line drawn through G and parallel to FE bisects DA.
Ans. (a)Give: A Q uadrilatreal ABCD in which AD=BC.PQ,R and S are mid-points of AB,BD, CD and AC respectively
To Prove: PQRS is a rhombus
Proof: InABD, P and Q are mid-points of AB and BD respectively PQ AD and PQ
1
... 1 2AB
Again in BCD R, and Q are mid-point of DC and BD respectively (given)
1 ... 3
2
1 , 2 3
,
hom PS BC and PS BC
AD BC
From and
PS RQ and PQ PS RQ
Now PS RQ and PS RQ PQRS is gm
Further PQ RS PS RQ PQRS is a r bus
Hence, the result.
(b)Given: ABCD is a kite in which BC=CD, AB = AD, E, F, G are mid-point of CD, BC and AB respectively.
To Prove: (i) EFG 90
(ii). The line draw through G and parallel to FE bisect DA.
Construction: Join AC and BD and Draw GH through G parallel to FE.
Proof: (i) Diagonal of kite intersect at right angle
90 .. 1
MON In BCD
E and F are mid-point of CD and BC respectively
1 ... 2
2
180
90 180
180 90 90
90 Pr
EF DB and EF DB EF DB MF ON
MON MFN MFN
MFN MFN
EFG oved
(ii). In ABD
G is mid-point of AB and HG DB
[From (2), EF DB and EF HG given
] HG DB
H is mid-point of DA
Hence, the draw through G and parallel FE bisects DA.
Q.12 In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid- point of CD. Calculate:
(i) BG if AD=6cm (ii) CF if GE = 2.3 cm (iii) AB if BC = 2.4cm (iv) ED if FD = 4.4cm.
Ans. Given: The straight line l, m, and n are parallel to each other. G is the mid-point of CD.
To Calculate: (i) BG if AD= 6cm
(ii). CF if GE = 2.3 cm (iii) AB if BC = 2.4 cm (iv) ED if FD = 4.4 cm (i) In ACD,
G is the mid-point of CD (given) and BG AD (m n given)
B is the mid-point of AC and 1 BG 2AD (ii) In CDF
G is the mid-point of CD and GF CF
E is mid-point of DF and 1 GF 2CF
2 2 2.3cm 4.6
CF GE cm
Ans.
(iii) Since B is mid-point of AC (In pert (i))
2.4 ( 2.4 (given))
AB BC
AB cm BC cm
(iv) Since E is mid-point of FD
1 2
1 4.4 (FD 4.4 (given))
2
ED 2.2 cm .
ED FD
ED cm cm
Ans
