Lesson: Quadrilaterals
Exercise 8.1 (12)
Question: 1
The angles of quadrilateral are in the ratio
3: 5: 9:13
.
Find all the angles of the quadrilateral.
Solution:
Let ‘
a’
be the common ratio between the angles.
Then the angles of the quadrilateral are
and 3 , 5 , 9 13 .a a a a
Sum of the interior angles of a quadrilateral
360
Then,
3 5 9 13 360
30 360
12
a a a a
a
a
Hence, the angles of the quadrilateral are:
3 3 12 36
5 5 12 60
9 9 12 108
a
a
a
and
13 13 12 156a
Question: 2
If the diagonals of a parallelogram are equal, then
show that it is a rectangle.
Solution:
Let
ABCD
be the parallelogram and
AC BD
.
We need to prove that
ABCD
is a rectangle.
In
ΔABC
and
ΔBAD
,
AB BA
(Common)
BC AD
(Opposite sides of a parallelogram are
equal)
AC BD
(Given)
Therefore,
ABC BAD
(By
SSS
congruence condition)
A B
(By CPCT)
Also,
180A B
(Sum of the angles on the
same side of the transversal)
2 180
90
A
A
Since
ABCD
is a parallelogram with one angle equal
to
90
ABCD
is a rectangle.
Hence, it is proved.
Question: 3
Show that if the diagonals of a quadrilateral bisect
each other at right angles, then it is a rhombus.
Solution:
Let
ABCD
be a quadrilateral whose diagonals
AC
and
BD
bisect each other at
O
.
Then,
,OA OC OB OD
and
90 .AOB BOC COD DOA
We need to show that ABCD is a rhombus.
AB BC CD AD
Now, in
AOB
and
,COB
OA OC
(Given)
AOB COB
90
(Given)
OB OB
(Common)
Therefore,
AOB COB
by
SAS
congruence condition.
Thus,
AB BC
(By CPCT)
Similarly, we can prove that
AB BC CD AD
Thus, ABCD is a rhombus.
Question: 4
Show that the diagonals of a square are equal and
bisect each other at right angles.
Solution:
Let
ABCD
be a square and its diagonals
AC
and
BD
intersect each other at
O
.
We need to show that the diagonals of a square
ABCD
are equal and bisect each other at right angles.
i.e.,
, ,AC BD AO OC OB OD
and
90AOB
Now, in
ABC
and
,BAD
BA AB
(Common)
90ABC BAD
BC AD
(
ABCD
is a square)
∴
ABC BAD
by
SAS
congruence rule.
Thus,
AC BD
by
CPCT
Hence, the diagonals are equal.
Now, in
AOB
and
COD
,
BAO DCO
(Alternate interior angles)
AOB COD
(Vertically opposite angles)
AB CD
(Sides of the square)
Therefore,
AOB COD
by AAS congruence condition.
Thus,
AO CO
and
OB OD
by CPCT. (Diagonal bisects
each other.)
Now, in
AOB
and
COB
,
OB OB
(Common)
AO CO
(As proved above)
AB CB
(Sides of the square)
Therefore,
AOB COB
by
SSS
congruence
condition.
AOB COB
But
180AOB COB
(Linear pair)
Thus,
90AOB COB
Similarly, we can show that,
90DOA COD
Hence, it is proved that diagonals bisect each other
at right angles.
Question: 5
Show that if the diagonals of a quadrilateral are
equal and bisect each other at right angles, then it is
a square.
Solution:
Let
ABCD
be a quadrilateral in which diagonal
AC
and
BD
are equal and bisect each other at right
angle
s
.
We need to prove that quadrilateral
ABCD
is a
square.
Now, In
AOB
and
COD
,
AO CO
(
Diagonals bisect each other)
AOB COD
(Right angles)
OB OD
(
Diagonals bisect each other)
Therefore,
AOB COD
by
SAS
congruence
condition.
Thus,
AB CD
(By
CPCT ...
….. (i)
Also,
OAB OCD
(By
CPCT
)
||AB CD
(
Alternate interiors angles are equal)
Similarly, we can prove that
.AOD COB
Thus,
AD BC
by CPCT …………. (ii)
And
||AD BC
Thus,
ABCD
is a parallelogram.
Now, in
AOD
and
COD
,
AO CO
(
Diagonals bisect each other)
AOD COD
(Right angles)
OD OD
(Common)
Therefore,
AOD COD
by SAS congruence rule.
Thus,
AD CD
(By CPCT) --- (iii)
Now using (i), (ii) and (iii), we get
AD BC CD AB
Hence,
ABCD
is parallelogram with all sides equal.
Now, let us consider
ΔACD
and
ΔBDC
.
AC BD
(Given)
DC DC
(Common)
AD BC
(As proved above)
So, by
SSS
congruence rule,
Δ ΔACD BDC
ADC BCD
by CPCT.
But,
180ADC BCD
(Co-interior angles)
2 180
90
ADC
ADC
Thus, ABCD is a parallelogram with one angle equal
to
90
.
ABCD
is a rectangle. But,
ABCD
has all equal
sides.
Hence, the given quadrilateral
ABCD
is a square.
Question: 6
Diagonal
AC
of a parallelogram
ABCD
bisects
A
(See Fig. 8.19).
Show that
(i) it bisects
C
also,
(ii)
ABCD
is a rhombus.
Solution:
(i)To show:
AC
bisects
C
.
AB DC
(Opposite sides of
gm
)
DAC ACB
and
CAB ACD
But,
DAC CAB
ACB ACD
AC
bisects
C
(ii) In
,ACD
ACD CAB
(Alternate interior angle)
But,
CAB CAD
(
AC
bisects angle
A
)
ACD CAD
AD CD
(Opposite sides of equal angles)
Also,
AB CD
and
BC DA
(Opposite sides of a
|| gm
)
So,
AB CD AD BC
Thus,
ABCD
is a rhombus.
Question: 7
ABCD is a rhombus. Show that diagonal
AC
bisects
A
as well as
C
and diagonal
BD
bisects
B
as
well as
D
.
Solution:
Since
ABCD
is a rhombus,
So,
AB BC CD DA
In ,ABC
AB BC
BCA BAC
… (i)
(Angles opposite of equal sides of a triangle are equal)
Also,
||AB CD
(
ABCD
is a rhombus)
DAC BCA
(Alternate interior angles)
BAC DAC
(From (i))
Therefore,
AC
bisects
.A
Similarly, the diagonal
AC
bisects
C
.
Also, by above method we can also prove that
diagonal
BD
bisects
B
as well as
.D
Hence, it is proved.
Question: 8
ABCD
is a rectangle in which diagonal
AC
bisects
A
as well as
C
. Show that:
(i)
ABCD
is a square
(ii) Diagonal
BD
bisects
B
as well as
D
.
Solution:
(i)
Since
ABCD
is a rectangle,
||AD BC
DCA BAC
(Alternate interior angle)
But,
DCA ACB
(
AC
bisects
C
)
BAC ACB
Now, in
,ABC
BAC ACB AB BC
(Sides opposite to equal
angles of a triangle are equal)
Also,
CD AB
(Opposite sides of a rectangle)
Therefore,
AB BC CD AD
.
Thus,
ABCD
is a square.
(ii)
In Δ
BCD
,
BC CD
(As proved above)
CDB CBD
(Angles opposite to equal sides
are equal).
Also,
CDB ABD
(Alternate interior angles)
CBD ABD
Thus,
BD
bisects
B
.
Now,
CBD ADB
(Alternate interior angle)
( )CDB ADB CDB CBD
Thus,
BD
bisects
D
.
Question: 9
In parallelogram
ABCD
, two points
P
and
Q
are
taken on diagonal
BD
such that
DP BQ
(See Fig.
8.20). Show that:
(i)
APD CQB
(ii)
AP CQ
(iii)
AQB CPD
(iv)
AQ CP
(v)
APCQ
is a parallelogram
ABQ CDP
(Alternate interior angles)
AB CD
(Opposite sides of a
|| gm
)
Thus,
AQB CPD
by
SAS
congruence rule.
(iv)
AQ CP
by
CPCT
(Using
AQB CPD
from part (iii))
(v) From (ii) and (iv),
APCQ
has opposite sides equal.
A quadrilateral is called a parallelogram if its
opposite sides are equal. Thus,
APCQ
is a
parallelogram.
Question: 10
ABCD
is a parallelogram and
AP
and
CQ
are
perpendiculars from vertices
A
and
C
on diagonal
BD
(see Fig. 8.21).
Show that
(i)
APB CQD
(ii)
AP CQ
Question: 11
In and ||, , ,ABC DEF AB DE AB DE BC EF
And
||BC EF
. Vertices
A, B
and
C
are joined to
vertices
D, E
and
F
respectively (see Fig. 8.22).
Show that
(i) Quadrilateral
ABED
is a parallelogram
(ii) Quadrilateral
BEFC
is a parallelogram
(iii)
||AD CF
and
AD CF
(iv) quadrilateral
ACFD
is a parallelogram
(v)
AC DF
(vi)
. ABC DEF
Solution:
(i) In quadrilateral
ABED,
AB DE
and
||AB DE
(Given)
So, the quadrilateral
ABED
is a parallelogram.
(ii) In quadrilateral
BEFC,
BC EF
and
||BC EF
(Given)
So, the quadrilateral
BEFC
is a parallelogram.
(iii) Since
ABED
is a parallelogram,
AD BE
and
...AD BE i
(Opposite sides of a parallelogram)
Also,
BEFC
is a parallelogram.
BE CF
and
...BE CF ii
(Opposite sides of a parallelogram)
From Eq. (i) and (ii),
AD CF
and
| .|AD CF
(iv) From part (iii), we have
AD CF
and
| .|AD CF
Thus,
ACFD
is a parallelogram.
(v) Since
ACFD
is a parallelogram (From part (iv)),
AD BC
(See Fig. 8.23).
Show that
(i)
A B
(ii)
C D
(iii)
ABC BAD
(iv)
Diagonal iagonal AC D BD
[Hint: Extend
AB
and draw a line through
C
parallel
to
DA
intersecting
AB
at
E.
]
Solution:
Let us draw a line through
C
parallel to
DA
intersecting
AB
at
E
when
AB
is extended.
(i) In
,ADCE
AD CE
(By construction)
and
AE DC
(∵
AB CD
is given).
Thus,
ADCE
is a parallelogram.
CE AD
(Opposite sides of a parallelogram)
But,
AD BC
(Given)
BC CE
So, in
∆BCE
,
CBE CEB
(Angle opposite to equal sides)
also,
180A CEB
.
(
ADCE
is a parallelogram)
⇒
180A CBE
…. (1) (
CBE CEB
)
but,
180ABC CBE
…(2) (Linear pair)
Comparing (1) & (2), we get
A ABC
(ii) In
,ABCD AB CD
(Given)
180A D B C
(Angles on the same sides of transversal)
( )A D A C A B
D C
(iii) InΔ
ABC
and
ΔBAD
,
AB BA
(Common)
DAB CBA
(As proved above in (i))
AD BC
(Given)
Thus,
ABC BAD
(By
SAS
congruence condition)
(iv)
AC BD
by
CPCT
as
.ABC BAD
Also,
AC
&
BD
are diagonals of
ABCD
.
Exercise 8.2 (7)
Question: 1
ABCD
is a quadrilateral in which
P
,
Q
,
R
and
S
are
midpoints of the sides
AB, BC, CD
and
DA
(See Fig
8.29).
AC
is a diagonal. Show that:
(i)
||SR AC
and
1
2
SR AC
(ii)
PQ SR
(iii)
PQRS
is a parallelogram.
Solution:
(i) In
ΔDAC
,
R
and
S
are the midpoints of
DC
and
DA
respectively.
Thus, by the midpoint theorem,
||SR AC
and
1
2
SR AC
(ii) In
ΔBAC,
P
and
Q
are the midpoints of
AB
and
BC
respectively.
Thus by the midpoint theorem,
||PQ AC
and
1
2
PQ AC
also,
1
2
SR AC
(Using part(i)).
Thus,
PQ SR
(iii)
||SR AC
from (i)
and,
||PQ AC
from (ii)
||SR PQ
from (i) and (ii)
also
,
PQ SR
(from part (ii))
Thus,
PQRS
is a parallelogram.
Question: 2
ABCD
is a rhombus and
P, Q, R
and
S
are the
midpoints of the sides
AB, BC, CD
and
DA
respectively. Show that the quadrilateral
PQRS
is a
rectangle.
Solution:
Let us join
AC
and
BD
.
In
ΔDRS
and
ΔBPQ,
DS BQ
1
(
1
2 2
)AD BC AD BC
SDR QBP
( )ADC CBA
1 1
2 2
( )DR BP DC AB DC AB
Thus,
DRS BPQ
by
SAS
congruence condition.
RS PQ
by
CPCT
--- (i)
Similarly,
QCR SAP
by
SAS
congruence condition.
RQ SP
by
CPCT
--- (ii)
Now, in
ΔABC,
P
and
Q
are the midpoints of
AB
and
BC
respectively.
So, by the midpoint theorem,
... | |PQ AC iii
Also, in
∆ADC,
R
and
S
are the midpoints of
DC
and
AD
respectively.
So, by the midpoint theorem,
||SR AC
||PQ SR
(Using (iii))
Thus,
PQRS
is a parallelogram.
Consider
,OMQN
( )OM QN BD QR
and
( )MQ ON PQ AC
OMQN
is a parallelogram.
MON MQN
(Opposite angles of a
parallelogram)
But,
90MON
(
Diagonals of a rhombus are perpendicular to each
other)
90
90
MQN
PQR
Hence,
PQRS
is a parallelogram with one right angle.
Thus,
PQRS
is a rectangle.
Question: 3
ABCD
is a rectangle and
P, Q, R
and
S
are midpoints
of the sides
AB, BC, CD
and
DA
respectively. Show
that the quadrilateral
PQRS
is a rhombus.
Solution:
Let us join
AC
and
BD.
We need to show that
PQRS
is a rhombus.
Now, in
ADB
,
P
and
S
are the midpoints of
AB
and
AD
respectively.
Thus,
||PS BD
and
...
1
2
PS BD i
(Using the midpoint theorem)
In
,BDC
R
and
Q
are the midpoints of
DC
and
BC
respectively.
||RQ BD
and
1
2
RQ BD
(Midpoint theorem) …(ii)
From Eq. (i) and (ii),
||PS RQ
and
PS RQ
Thus,
PQRS
is a parallelogram.
||PQ SR
and
PQ SR
(Opposite sides of parallelogram) --- (iii)
Now, in
ABC
,
P
and
Q
are the midpoints of side
AB
and
BC
respectively.
Thus,
||PQ AC
and
1
2
PQ AC
(Midpoint theorem) --
- (iv)
Now,
AC BD
(Diagonals of a rectangle are equal) -
-- (v)
From equations (i), (ii), (iii), (iv) and (v), we get
PQ QR SR PS
So,
PQRS
is a rhombus.
Question: 4
ABCD
is a trapezium in which
,||AB DC BD
is a
diagonal and
E
is the midpoint of
AD
. A line is
drawn through
E
parallel to
AB
intersecting
BC
at
F
(See Fig. 8.30). Show that
F
is the midpoint of
BC
.
Solution:
Let
EF
intersects
BD
at
G
.
Now, in
ΔBAD
,
E
is the midpoint of
AD
(Given)
and
|| ( .)EG AB EF AB
So,
G
is the midpoint of
BD
(Converse of the
midpoint theorem).
Now, in
ΔBDC
,
G
is the midpoint of
BD
(As proved above)
and
|| ( )GF DC EF AB DC
Hence,
F
is the midpoint of
BC
(Converse of the
midpoint theorem).
Question: 5
In a parallelogram
ABCD
,
E
and
F
are the midpoints
of sides
AB
and
CD
respectively (See Fig. 8.31). Show
that the line segments
AF
and
EC
trisect the
diagonal
BD
.
Solution:
E
and
F
are the midpoints of sides
AB
and
CD
respectively in a parallelogram
ABCD
.
We need to show that
.DP PQ QB
Since in the quadrilateral
AECF
,
|| ( )AE FC AB CD
and
)(
1 1
2 2
AE FC AB CD AB CD AE FC
Thus,
AECF
is a parallelogram.
||AF EC
and
AF EC
(Opposite sides of a
parallelogram).
Now, in
ΔDQC
,
F
is midpoint of the side
DC (
Given)
and
.|| ( || )FP CQ AF EC
P
is the midpoint of
DQ
(Converse of the midpoint
theorem)
DP PQ i
Similarly, in ∆
APB
,
E
is midpoint of side
AB
(Given)
and
.|| || ) (EQ AP AF EC
Q
is the midpoint of
PB
(Converse of the midpoint
theorem)
PQ QB ii
From equations (i) and (ii),
DP PQ BQ
Hence, the line segments
AF
and
EC
trisect the
diagonal
BD
.
Question: 6
Show that the line segments joining the mid-points
of the opposite sides of a quadrilateral bisect each
other.
Solution:
Let
PQRS
be a quadrilateral and
A, B, C
and
D
are
the midpoints of
PQ, QR, RS
and
SP
respectively.
We need to show that
AC
and
BD
bisect each other.
Now, in
ΔPSQ,
D
and
A
are the midpoints of
PS
and
PQ
respectively.
Thus,
1||DA SQ
(Using the midpoint theorem)
Similarly, in
ΔSRQ,
C
and
B
are the midpoints of
SR
and
RQ
respectively.
Thus,
| 2|BC SQ
From (1) and (2),
||DA BC
Similarly, we can show that
||DC AB
.
Thus,
ABCD
is parallelogram.
AC
and
BD
are the diagonals of the parallelogram
ABCD
. So, they will bisect each other.
Question: 7
ABC
is a triangle right angled at
C
.
A
line through
the midpoint
M
of hypotenuse
AB
and parallel to
BC
intersects
AC
at
D
. Show that
(i)
D
is the midpoint of
AC
(ii)
MD AC
(iii)
1
2
CM MA AB
Solution:
(i) In
ΔACB
,
90C
.
Since
M
is the midpoint of
AB
and
||MD BC
,
by the converse of the midpoint theorem,
D
is the midpoint of
AC.
CD DA
(ii)
||MD BC
90ADM ACB
(Corresponding angles).
Hence,
.MD AC
(iii) In
ΔAMD
and
ΔCMD,
AD CD
(As proved above)
ADM CDM
(Each 90°)
DM = DM
(Common)
Thus,
AMD CMD
(By
SAS
congruence
condition)
AM CM
by
CPCT
Also,
1
2
AM AB
(
M
is midpoint of
AB
)
Hence,
1
2
CM MA AB
.