Lesson: Triangles
Exercise 7.1
Question: 1
Which of the following is not a criterion for
congruence of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Solution:
c
The rule of congruence of triangles are SSS
(Side-Side-Side), SAS (Side-Angle-Side), ASA
(Angle-Side-Angle), AAS and RHS (Right
angle-Hypotenuse-Side). So, SSA is not a criterion for
congruence of triangles.
Question: 2
If
,AB QR BC PR
and
CA PQ
, then:
(a)
ABC PQR
(b)
CBA PRQ
(c)
BAC RPQ
(d)
PQR BCA
Solution:
b
Since,
,AB QR BC RP
and
CA PQ
A corresponds to Q, B corresponds to R &C
corresponds to P.
i.e.,
ABC QRP
which can also be written as
.CBA PRQ
Question: 3
In
and
50 .B
Then
C
is
equal to:
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Solution:
b
Given: In
,ABC AB AC
and
50B
To find:
C
In
, .ABC AB AC
B C
(
Angles opposite to equal sides are equal).
Since
50B
(Given)
50C
Question: 4
In
,ABC BC AB
and
80 .B
Then
A
is
equal to:
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Solution:
c
Given: In
,ABC BC AB
and
80 .B
To find:
A
In
,ABC BC AB
. iC A
( Angles opposite to equal sides
are equal)
Since, the sum of all the angles of a triangle is 180°
180
80 180 80
A B C
A C B
80 180A A
(Using eqn. (i))
2 180 80 100A
100
2
A
50A
Question: 5
In
,PQR R P
and
4 cmQR
and
5 cm.PR
Then the length of PQ is:
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution:
a
Given:
, , 4 cmPOR R P QR
and
5 cm.PR
In
, .PQR R P
PQ QR
(
Sides opposite to equal angles of a
triangle are equal)
Given
4 cm
4 cm
PQ QR
PQ
Hence, the length of PQ is 4 cm.
Question: 6
D is a point on the side BC of
ABC
bisects
.BAC
Then,
(a)
BD CD
(b)
BA BD
(c)
BD BA
(d)
CD CA
Solution:
b
Given: In
bisects
BAC
and D is a point
on side BC.
In
is an exterior angle.
[
Exterior angle > interior opposites angle]
Using (i) & (ii), we have
BA BD
(In a triangle, side opposite to the greater
angle is greater)
Similarly,
In
is an exterior angle.
[
Exterior angle > interior
opposites angle]
Using (i) & (iii), we have
AC CD
(In a triangle, side opposite to the greater
angle is greater).
Hence, using both the results we conclude that, option
(a), (c) & (d) cannot be true. Hence, the correct answer
is option (b).
Question: 7
It is given that
ABC FDE
and
5 cm,AB
40B
and
80 .A
Which one of the following
is true?
a 5 cm, 60°
b 5 cm, 60°
c 5 cm, 60°
d 5 cm, 0
°
4
DF F
DF E
DE E
DE D
Solution:
b
Given:
ABC FDE
and
5 cmAB ,
40B
and
80A
In ABC,
80A
and
40B
Since, the sum of all the angles of a triangle is 180°
180
80 40 180
180 120
60 i
A B C
C
C
C

Since,
ABC FDE
AB FD
[By CPCT]
5 cm FD
[Given
5 cmAB
]
We can write
5 cmFD
And also,
C E
[By CPCT]
60 E
[Using (i)]
Question: 8
Two sides of a triangle are of lengths 5 cm and 1.5 cm.
The length of the third side of the triangle cannot be:
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution:
d
Let us consider, ABC with sides
5 cmBC
and
1.5 cm.CA
In a triangle, the difference of two sides < third side
and sum of two sides > third side.
BC CA AB
and
BC CA AB
5 1.5 AB
and
5 1.5 AB
3.5 AB
and
6.5 AB
Here, all the options except (d) satisfy the inequality.
Question: 9
In
,PQR
if
R Q
, then:
a
b
c
d
QR PR
PQ PR
PQ PR
QR PR
Solution:
b
In
, .PQR R Q
PQ PR
[Side opposite to greater angle is longer]
Question: 10
In triangles ABC and PQR,
,AB AC C P
and
.B Q
The two triangles are:
(a) Isosceles but not congruent
(b) Isosceles and congruent
(c) Congruent but not isosceles
(d) Neither congruent nor isosceles
Solution:
a
Given, In
, .ABC AB AC
. i C B
[Angles opposite to equal sides
are equal]
So, ABC is an isosceles triangle.
Now, it is also given that
C P
and
.B Q
P Q
[Using eq. (i)]
Now, In
,PQR
P Q
[Proved above]
QR PR
[Sides opposite to equal angles are equal]
So,
PQR
is an isosceles triangle.
Therefore, both triangles are isosceles.
But, corresponding sides of the triangles are not equal
and AAA is not a criterion for the congruence of
triangles. Hence, the triangles are not congruent.
Question: 11
In triangles ABC and
,DEF AB FD
and
.A D
The two triangles will be congruent by SAS axiom if:
a
b
c
d
BC EF
AC DE
AC EF
BC DE
Solution:
b
Given, triangles
ABC
and
,DEF AB FD
and
.A D
Two triangles are congruent by SAS axiom if two
corresponding sides and the included angle are equal.
i.e.,
ABC DEF
Therefore, if
AC DE
, the triangles are congruent by
SAS axiom.
Exercise 7.2
Question: 1
In triangles ABC and
,PQR A Q
and
.B R
Which side of PQR should be equal to side AB of
ABC so that the two triangles are congruent?
Solution:
Given,
∆ABC & ∆PQR,
A Q
and
.B R
AB and QR are the included sides.
So, if
AB QR
, then
ABC QRP
(By ASA axiom).
Question: 2
In triangles ABC and PQR,
A Q
and
.B R
Which side of PQR should be equal to side BC of
ABC so that the two triangles are congruent?
Solution:
Given, in triangles ABC and PQR,
A Q
and
.B R
RP is corresponding side to BC.
So, if
BC RP
, then
ABC QRP
(By AAS axiom).
Question: 3
“If two sides and an angle of one triangle are equal to
two sides and an angle of another triangle, then the
two triangles must be congruent. Is the statement true?
Why?
Solution:
No; because with respect to congruency rule, two
sides and the included angle of one triangle are equal
to two sides and the included angle of another triangle,
which is called SAS rule.
Question: 4
“If two angles and a side of one triangle are equal to
two angles and a side of another triangle, then the two
triangles must be congruent.” Is the statement true?
Why?
Solution:
No, because side of one triangle must be equal to the
corresponding side of the other triangle.
Question: 5
Is it possible to construct a triangle with sides
measuring 4 cm, 3 cm and 7 cm? Give reason for your
Solution:
No; because in a triangle the sum of any two sides is
greater than the third side, but
4 3 7.
Question: 6
It is given that
.ABC RPQ
Is it true to say that
BC QR
? Why?
Solution:
No; because as per the congruency rule, sides &
angles of one triangle should be equal to the
corresponding sides and angle of the other triangle.
Here,
ABC RPQ
, , AB RP BC PQ AC RQ
[By CPCT]
Hence, BC is not necessarily equal to QR.
Question: 7
If
PQR EDF
, then is it true to say that
PR EF
?
Solution:
Yes; because as per the congruency rule, sides &
angles of one triangle should be equal to the
corresponding sides and angle of the other triangle. If
,PQR EDF
, , PQ ED QR DF RP FE
[By CPCT]
Question: 8
In
, 70PQR P
and
30 .R
Which side of this
Solution:
In
,PQR
70 , 30
180
70 30 180
180 70 30
180 100
80
P R
P Q R
Q
Q
Therefore Q is largest angle.
This means PR is the longest. (Side opposite to the
largest angle is longest)
Question: 9
AD is a median of the triangle ABC. Is it true that
Solution:
In triangle ABD,
(Sum of any two sides of a triangle is greater than the
third side)
In triangle ACD,
(Sum of any two sides of a triangle is greater than the
third side)
2
2
2
AB BC AC AD BD CD BC
Hence, it is proved.
Question: 10
M is a point on side BC of a triangle ABC such that
AM is the bisector of
.BAC
Is it true to say that
perimeter of the triangle is greater than 2AM? Give
Solution:
In triangle ABC, AM is the bisector of angle BAC.
In triangle ABM. we have
.. 1AB BM AM
[Sum of any two sides of a
triangle is always greater than its third side]
Similarly, in triangle AMC, we have
AC CM AM
[Sum of any two sides of a triangle
is always greater than its third side] (2)
On adding (1) and (2) we get,
( ) 2
2
AB BM AC CM AM AM
AB BM CM AC AM
AB BC AC AM
Hence, it is proved
Question: 11
Is it possible to construct a triangle with sides
measuring 9 cm, 7 cm and 17 cm? Give reason for
Solution:
No. Because, in any triangle, the sum of any two sides
should be greater than the third side.
But in this case,
9 7 16 17.
Question: 12
Is it possible to construct a triangle with sides
measuring8 cm, 7 cm and4 cm? Give reason for your
Solution:
Yes, because in each case the sum of two sides is
greater than the third side.
. ., 7 4 8,
8 4 7,
7 8 4.
i e
Exercise 7.3
Question: 1
ABC is an isosceles triangle with
AB AC
and
BD
and
CE
are its two medians. Show that
BD CE
.
Solution:
Given:
,AB AC BD
and CE are two medians.
In
ABD
and
,ACE
A A
Common
AB AC
Given
(
D
is midpoint of
,AC
and
E
is midpoint of
.AB
)
ABD ACE
(By SAS property)
BD CE
(By CPCT)
Question: 2
In Fig.7.4, D and E are points on side BC of a ABC
such that
BD CE
and
Show that
.ABD ACE
Solution:
Given: In ABC,
BD EBC
and
(Given)
2 1
(Angles opposite to equal sides of a triangle
are equal.)
1 3 180 ................(i)
2 4 180 ...............(ii)
[Linear pair axiom]
From (i) and (ii)
1 3 2 4
1 2
3 4
 
In ABD and ACE
(Given)
3 4
(As proved above)
BD CE
(Given)
Δ ΔABD ACE
(By
SAS
property)
Question: 3
CDE is an equilateral triangle formed on a side CD of
a square ABCD (Fig. 7.5). Show that
Solution:
In
,DEC DE CE
(All sides of equilateral are
equal)
...(i)4 3
[Angles opposite to equal sides of a
triangle are equal.]
And,
2 1 90
(ii)
From (i) and (ii),
4 2 1 3  
(Sides of square)
by SAS property.
Question 4
In Fig. 7.6,
,BA AC DE DF
such that
BA DE
and
.BF EC
Show that
.ABC DEF
Solution:
In
ABC
and
DEF
90A D
(Given)
BF CE
(Sides opposite to equal angles are equal)
BF CF CE CF
BC EF
AB DE
(Given)
ABC DEF
(By RHS property)
Question: 5
Q is a point on the side SR of a PSR such
that
PQ PR
. Prove that
.PS PQ
Solution:
In
,PRQ
PR PQ
1 R  
(Angles opposite to equal sides of a
triangle are equal.)
But,
1 S
(
1
is an exterior angle for
PQS
)
( 1 )R S R
PS PR
(
In any triangle, side opposite to larger
angle is longer)
)( PR PQPS PQ
Hence, the result.
Question: 6
S is any point on side QR of PQR. Show that
2 .PQ QR RP PS
Solution:
In PQS,
... iPQ QS PS
(
Sum of two sides of a is
greater than the third side)
In PSR,
i( i)RS PR PS
(
Sum of two sides of a
is greater than the third side)
Adding eq. (i) and eq. (ii), we get
2
2 ( ).
PQ QS SR PR PS
PQ QR PR PS QS SR QR
Question: 7
D is any point on side AC of a ABC with
.AB AC
Show that
.CD BD
Solution:
In ABC, Join BD
AB AC
(Given)
C ABC
(Angles opposite to equal sides of a
triangle are equal.)
But,
) (CBD ABCAB A CBDC BD
C CBD
(
C ABC
, as proved above)
Then, in DBC,
BD CD
(As the side opposite to greater angle is longer.)
Question: 8
In Fig. 7.7, l ||m and M is the midpoint of a line
segment AB. Show that M is also the midpoint of any
line segment CD, having its end points on l and m,
respectively.
Solution:
Given: l
m, M is midpoint of line segment AB, i.e.,
AM MB
We need to prove that
.MC MD
Let us consider, AMC and BMD.
Since,
l m
and AB is the transversal,
1 2
(Alternate interior angles)
AM BM
(Given)
3 4
(Vertical opposite angles)
AMC BMD Δ
by ASA property
MC MD
(By CPCT)
Question 9
Bisectors of the angles B and C of an isosceles triangle
with
AB AC
intersect each other at O. BO is
produced to a point M. Prove that
.MOC ABC
Solution:
In
OB
and
OC
are the angle bisectors of
B
and
C
respectively.
In
,ABC
AB AC
(Given)
ACB ABC
(Angles opposite to equal sides of
a triangle are equal.)
1 1
2 2
ACB ABC
2 1... i
bisects
bisects
BO ABC
CO ACB
1 2MOC
(Exterior angle property of
triangle.)
2 1MOC
(Using (i))
( 2 1)MOC ABC ABC
Hence, it is proved.
Question 10
Bisectors of the angles B and C of an isosceles triangle
ABC, with
,AB AC
intersect each other at O. Show
that the external angle adjacent to
ABC
is equal to
BOC
Solution:
In
,ABC AB AC
(Given)
ACB ABC
(Angles opposite to equal sides of
a triangle are equal.)
1 1
2 2
ACB ABC
2 1
(
BO
bisects
,ABC CO
bisects
ACB
)
In
,BOC
1 2 180BOC
(Angle sum property of
triangle.)
1 1 180BOC
(
1 2,
as proved
above)
2 1 180BOC
...180 (i)ABC BOC
(
BO
bisects
ABC
)
But,
....180 (ii)ABC DBA
(
DBC is a line segment)
On comparing eqn. (i) and (ii), we get,
DBA BOC
Question 11
In Fig. 7.8, AD is the bisector of
BAC
. Prove
that
.AB BD
Solution:
In ABC, AD is angle bisector of
.BAC
BAC
1 2
But,
[
is an exterior angle to
AB BD
(Side opposite to the larger angle is longer)
Hence, it is proved.
Exercise 7.4
Question: 1
Find all the angles of an equilateral triangle.
Solution:
Given,
,ABC AB AC
is an equilateral triangle.
ACB ABC
(Angle opposite to equal sides of a
triangle is equal.)
1 1
2 2
ACB ABC
2 1
(
BO
bisects
,ABC CO
bisects
ACB
)
In
,BOC
1 2 180BOC
(Angle sum property of a
triangle)
1 1 180BOC
(
1 2,
as proved
above)
2 1 180BOC
...180 (i)ABC BOC
(
BO
bisects
ABC
)
But,
....180 (ii)ABC DBA
(
DBC
is a line segment)
On comparing equations (i) and (ii), we get,
DBA BOC
Question: 2
The image of an object placed at a point A before a
plane mirror LM is seen at the point B by an observer
at D as shown in Fig. 7.12. Prove that the image is as
far behind the mirror as the object is in front of the
mirror.
[Hint: CN is perpendicular to the mirror.
Also, angle of incidence
angle of reflection].
Solution:
To prove:
AE BE
Proof:
&CN LM AB LM
AB NC
... iA i
(Alternate interior angles)
... iiB r
(Corresponding angles)
... iiii r 
(Incident angle
reflected angle)
Using (i), (ii) and (iii), we get,
A B
In
CEB
and
CEA
B A
(As proved above)
( )
0
1 2 90 AB LM
CE CE
(Common)
CEB CEA
by AAS rule
BE AE
(By CPCT)
Question: 3
ABC is an isosceles triangle with
AB AC
and D is
a point on BC such that
(Fig. 7.13). To
prove that
a student proceeded as
follows:
In
ABD
and
,ACD
AB AC
(Given)
B C
(Because
AB AC
) and
Therefore,
ABD ACD
(AAS)
So,
(CPCT)
What is the error in the above proof?
(Hint: Recall how
B C
is proved when
AB AC
).
Solution:
AB AC
cannot conclude that
ABD ACD
.
Here is the correct proof.
In
and
( )
0
AB AC
(Given)
(Common)
(By RHS)
(By CPCT)
Question: 4
P is a point on the bisector of
ABC
.
If a line through P, parallel to BA, meet BC at Q,
prove that BPQ is an isosceles triangle.
Solution:
2 3 ... i
(
BX
bisects
ABC
)
BA QP
(Given)
3 1 ... ii
(Alternate interior angles)
From (i) and (ii)
1 2
In
,BPQ
BQ PQ
(Converse of isosceles
property)
BPQ
is an isosceles triangle.
Question: 5
ABCD is a quadrilateral in which
AB BC
and
Show that BD bisects both the angles ABC
Solution:
In
DAB
and
.DCB
AB CB
(Given)
(Given)
DB DB
(Common)
DAB
DCB
by SSS property
1 2
3 4
DB
bisects
D
and
B
Question: 6
ABC is a right triangle with
.AB AC
Bisector of
A
meets
BC
at D. Prove that
Solution:
In right angle
,ABC
AB AC
Also,
BC
is the hypotenuse (
Hypotenuse longest
side)
90BAC
In
and
AC AB
(Given)
1 2
(
is an angle bisector of
A
)
(Common)
by SAS property
CD BD
(By CPCT)
and
....(i)ACD ABD
(By CPCT)
In ABC,
180A B C
90 180B B
(
90A
and using (i))
2 180 90 90B
45 45B C
(Using (i))
Also,
1 45 
and
2 45
(
bisects
A
and
90A
)
1 C  
and
2 B
In
Similarly, In
(Sides opposite to
equal angles are equal)
Hence,
(Using
above equations)
Therefore,
Question: 7
O is a point in the interior of a square ABCD such that
OAB is an equilateral triangle. Show that
OCD
is
an isosceles triangle.
Solution:
1 2 90 ..... i
(
ABCD
is a square)
3 4 60 ...... ii
(
OAB
is an equilateral
triangle)
Using (i) and (ii)
1 3 2 4 90 60
5 6 30
In
DAO
and
CBO
(Given)
5 6
(As proved above)
AO BO
(Given)
DAO CB
by SAS property
OD OC
OCD
is an isosceles triangle.
Question: 8
ABC and DBC are two triangles on the same base BC
such that A and D lie on the opposite sides of BC,
AB=AC and DB=DC. Show that AD is the
perpendicular bisector of BC.
Solution:
In triangle ABD and ACD
AB AC
(Given)
DB DC
(Given)
(Common)
Therefore,
ABD
is congruent to
.ACD
(By SSS
congruency rule)
So,
By CPCT1 2 
Now, in triangle AOB and AOC
AB = AC (Given)
Proved above1 2 
AO AO
(Common side)
Therefore,
AOB
is congruent to
AOC
So,
By CPCT3 4
and
BO OC
(By CPCT)
Since
3 4
Therefore,
3 4 90
Hence, proved that AD is the perpendicular bisector of
BC.
Question: 9
ABC
is an isosceles triangle in which
and
BE
are two Altitudes on sides
BC
and
AC
respectively. Prove that
.AE BD
Solution:
In triangles AEB and BDA,
1 2 90
In
,ABC
AC BC
[Given]
4 3
[By isosceles triangle property]
AB BA
(Common)
AEB BDA
(By AAS rule)
AE BD
(By CPCT)
Question: 10
Prove that sum of any two sides of a triangle is greater
than twice the median.
Solution:
Let us consider a ABC such that AD is a median on
BC i.e.
.BD BC
We need to prove that
Let’s produce AD to E such that
and join EC.
Now, in
and
EDC
(By construction)
1 2
(Vertically opposite angle)
DB DC
(Given)
By SAS property
So,
AB EC
(By CPCT)
Now, In
,AEC
AC CE AE
(
AB CE
as proved above
and
by construction)
Hence, it is proved.
Question: 11
, 2 .ABCD AB BC CD DA BD AC
Solution:
In
,AOB
iAO OB AB
(Sum of two sides is greater
than the third side in a triangle)
In
,BOC
ii OB OC BC
(Sum of two sides is greater than
the third side in a triangle)
In
,COD
iiiOC OD CD
(Sum of two sides is greater than
the third side in a triangle)
In
,DOA
(Sum of two sides is greater
than the third side in a triangle)
Adding (i), (ii), (iii) and (iv)

2
( )
2
AO OB OB OC OC OD OD OA
AB BC CD DA
AO OC BO OD AB BC CD DA
AC BD AB BC CD DA
Hence, it is proved.
Question: 12
Show that in a quadrilateral ABCD,
AB+BC+CD+DA>AC+BD.
Solution:
AB BC CD DA AC BD
In
,ABC
iAB BC AC
(Sum of two sides is greater
than the third side in a triangle)
In
,BCD
iiBC CD BD
(Sum of two sides is greater
than the third side in a triangle)
In
,CDA
iiiCD DA AC
(Sum of two sides is greater
than the third side in a triangle)
In
,DAB
(Sum of two sides is greater
than the third side in a triangle)
Adding (i), (ii), (iii) and (iv)
( )
( )
( ) ( )2 2
AB BC BC CD CD DA AD AB
AC BD AC BD
AB BC CD DA AC BD
AB BC CD DA AC BD
Hence, it is proved.
Question: 13
In a triangle ABC, D is the mid-point of side AC such
that
1
.
2
BD AC
Show that
ABC
is a right angle.
Solution:
To prove:
90ABC
1
... i
2
BD AC
(Given)
and
1
..... ii
2
(
D
is midpoint of
AC
)
From (i) and (ii)
In
DAB
1 A  
(Angle opposite equal sides are
equal) ... …(iii)
In
Δ ,DBC
BD CD
2C
(Angle opposite equal sides are
equal) ... ….(iv)
In
,ABC
180A ABC C
(Angle sum property of
triangle)
1 2 180ABC
1, 2A C
1 2 180ABC  
180ABC ABC
( 1 2 )ABC
2 180
90
ABC
ABC
Hence, it is proved.
Question: 14
In a right triangle, prove that the line-segment joining
the mid-point of the hypotenuse to the opposite vertex
is half the hypotenuse.
Solution:
Let us consider a right
ABC
with
90ABC
and let E be the midpoint of hypotenuse AC.
We need to prove
1
2
BE AC
.
Producing
BE
to
D
s.t.
ED BE
Join A to D and C to D
Since,
=ED EB
(By construction)
and
EA EC
(Given)
ABCD
is a parallelogram
90ABC
Also, in a parallelogram, if one angle is right, then all
angle are right.
Parallelogram
ABCD
rectangle
AC BD
(Diagonals of a rectangle are equal)
1 1
2 2
AC BD
1 1
2 2
AC BE BE ED BD
Hence, it is proved.
Question: 15
Two lines l and m intersect at the point O and P is a
point on a line n passing through the point O such that
P is equidistant from l and m. Prove that n is the
bisector of the angle formed by l and m.
Solution:
We draw
PQ l
and
.PR m
Now, in
OQP
and
,ORP
1 2 90
(By construction)
OP OP
(Common)
PQ PR
(Given P is equidistant from l and m)
OQP ORP
By RHS property
By CPCT
3 4
n
is the bisector of
QOR
Question: 16
Line segment joining the mid-points M and N of
parallel sides AB and DC, respectively of a trapezium
ABCD is perpendicular to both the sides AB and DC.
Prove that
.
Solution:
Joining AN and BN,
in
AMN
and
,BMN
AM BM
(Given)
1 2 90
NM NM
(Common)
AMN BMN
by SAS property
3 4
and,
...(i)AN BN
(By CPCT)
Now,
90DNM CNM
5 3 6 4
....... (5 6 (ii) 3 4)
Now, In
and
,BCN
DN CN
(Given)
5 6
(Using (ii))
AN BN
(Using (i))
by SAS property
(By CPCT)
Question: 17
ABCD is a quadrilateral such that diagonal AC bisects
the angles A and C. Prove that
and
.CB CD
Solution:
Given: In quadrilateral ABCD, diagonal AC bisects
A
and
.C
i.e.,
1 2
and
3 4
Now, in
ABC
and
1 2
(Given)
AC AC
3 4
(Given)
by ASA Property
(By CPCT)
CB CD
(By CPCT)
Question: 18
If ABC is a right angled triangle such that
AB AC
and bisector of angle C intersects the side AB at D.
Prove that
Solution:
Given: In figure ABC is a right angle,
AB AC
, CD
is bisector of
C
Construction: Draw
DE BC
In right angle
,ABC
AB AC
BC is the hypotenuse
90A
In
DAC
and
DEC
3 90A
1 2
(Given)
DC DC
(Common)
DAC DEC
by AAS rule.
... iDA DE
(CPCT)
... iiCA CE
(CPCT)
In
BAC
AB AC
C B
(Angle opposite equal sides are equal)
Now,
180A B C
90 2 180
2 90
45
B B C
B
B
In
BED
5 180 4B  
180 45 90
180 135
45
5B
... iiiDE BE
(Sides opposite equal angles are equal)
From (i) and (iii)
... ivDA DE BE
Now,
BC CE BE
CA DA
(Using (ii) and (iii)
Hence, it is proved.
Question: 19
AB and CD are the smallest and largest sides of a
B
and
D
decide
which is greater.
Solution:
AB is the shortest and CD is the longest side of ABCD.
We need to prove
B D
or
D B
.
Joining BD.
Now, in
,ABD
1 3... i
(Angle opposite to the longer side is greater)
In
BCD
CD BC
2 4...... ii
2 4...... ii
(Angle opposite to the longer side is greater)
Adding equation (i) and (ii), we get
1 2 3 4
B D
Question: 20
Prove that in a triangle, other than in an equilateral
triangle, angle opposite the longest side is greater than
2
3
of a right angle.
Solution:
Let us consider
ABC
with BC as the longest side.
We need to prove
2
90
3
BAC
In
,ABC
BC AB
(As BC is the longest side)
..... iA C
(In a
Δ
angle opposite the longer side is greater)
Similarly,
BC AC
(As BC is the longest side)
... iiBA
Adding (i) and (ii) we get,
2 A B C
A
on both the sides,
0
3
3 180
180
3
2
90
3
A A B C
A
A
2
3
A
right angle.
Question: 21
Prove that AC is the perpendicular bisector of BD.
Solution:
In
ABCD
,
and
CB CD
We Join AC and BD.
Let AC and BD intersect at point O.
In
ABC
and
(Given)
BC CD
AC AC
by SSS property
1 2
(CPCT)
In
AOB
and
AOD
(Given)
1 2
(As proved above)
AO AO
AOB AOD
by SAS property
BO DO
(CPCT)
3 4
But,
3 4 180
(Linear pair axiom)
2 3 180
3 90
4 90
Hence,
AC
is perpendicular bisector of
3 90
.B
BO DO