6 TRIANGLES CHAPTER

CHAPTER
6
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TRIANGLES
Points to Remember :
1. Two figures having the same shape but not necessarily the same size are called similar figures.
2. All the congruent figures are similar but the converse is not true.
3. Two polygons having the same number of sides are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are proportional (i.e. in the same ratio)
4. Basic proportionality theorem (Thales theorem) : If a line is drawn parallel to one side of a triangle to
intersect the other two sides in distinct points, then the others two sides are divided in the same ratio.
5. Converse of Thales’ theorem : If a line divides any two sides of a triangle in the same ratio, then the line
is parallel to the third side of the triangle.
6. The line drawn from the mid-point of one side of a triangle is parallel to another side bisects the third
side.
7. The line joining the mid-points of two sides of a triangle is parallel to the third side.
8. AAA similarity criterion : If in two triangles, corresponding angles are equal, then the triangles are
similar.
9. AA similarity criterion : If in two triangles, two angles of one triangle are respectively equal to the two
angles of the other triangle, then the two triangles are similar.
10. SSS similarity criterion : If in two triangles, corresponding sides are in the same ratio, then the two
triangles are similar.
11. SAS similarity criterion : If one angle of a triangle is equal to one angle of another triangle and the
sides including these angles are in the same ratio, then the triangles are similar.
12. The ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.
13. If the areas of two similar triangles are equal, then the triangles are congruent.
14. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then
the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.
15. Pythagoras Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the
squares of the other two sides.
16. Converse of Pythagoras Theorem : If in a triangle, square of one side is equal to the sum of the squares
of the other two sides, then the angle opposite to first side is a right angle.
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ILLUSTRATIVE EXAMPLES
Example 1. In the given figure, DE || OQ and DF || OR. Prove that EF || QR.
Solution.
[NCERT]
In POQ, DE || OR, by Thales’ theorem,
We have
MATHEMATICS–X
PE PD
=
EQ DO
...(1)
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Again, In POR, DF || OR, by Thales’ theorem,
P
PF PD
=
We have,
FR DO
from (1) and (2), we get
...(2)
D
E
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F
O
PE PF
Q
R
=
EQ FR

EF || QR
( converse of Thales’s theorem)
Example 2. Prove that the diagonals of a trapezium divide each other proportionally.
OR
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ABCD is a trapezium such that AB || DC. The diagonals AC and BD intersect at O. Prove that
Solution.
AO CO .
[CBSE 2004, NCERT]
=
BO DO
Through O, draw OE || CD (or AB). In ABD, EO || AB, by Thales’ theorem,
We have
AE BO
=
ED DO
A
...(1)
B
Again, In ADC, EO || DC, by Thales’ theorem,
We have
AE AO
=
ED OC
...(2)
from (1) and (2), we get
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BO AO
AO CO
=

=
DO OC
BO DO
E
B
O
D
Hence proved.
C
Example 3. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.
[CBSE 2005]
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OR
The diagonals of a quadrilateral ABCD intersect each other at the point O such that
Show that ABCD is a trapezium.
Solution.
AO CO
=
.
BO DO
[NCERT]
Through O, draw OE parallel to AB intersecting BC at E.
Now, in ABC, OE || AB

AO BE
=
OC EC
D
( by construction)
C
O
E
...(1) ( Thales’ theorem)
A
B
AO CO
AO BO
=
=
Now,

...(2) ( given)
BO DO
OC DO
from (1) and (2), we get
BE BO

EC DO
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MATHEMATICS–X
Now, In ABC, we have

BO BE
=
DO CE
(proved above)
OE || DC
( Converse of Thales’ theorem)
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Now, In quadrilateral ABCD, OE || AB and OE || DC  AB || DC.

Quadrilateral ABCD is a trapezium.
Example 4. In the given figure, P is the mid-point of BC and Q is the mid point of AP. If BQ when produced
meets AC at R, prove that RA =
Solution.
1
CA .
3
[CBSE 2006 C]
Draw PS || BR, meeting AC at S.
In BCR, P is the mid-point of BC and PS || BR.
 S is the mid-point of CR.
 CS = SR
...(1)
In APS, Q is the mid-point of AP and QR || PS.
 R is the mid-point of AS.
 AR = RS
...(2)
from (1) and (2), we get
AR = RS = SC
Now, AC = AR + RS + SC= 3AR
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A
R
Q
S
B
B
C
P
1
AR= CA
Hence proved.
3
Example 5. In the given figure, altitudes AD and CE of ABC intersect each other at the point P. Show that:
(i) AEP ~ CDP
(ii) ABD ~ CBE
(iii) AEP ~ ADB
(iv) PDC ~ BEC
[NCERT]
Solution. (i) In AEP and CDP, we have
AEP = CDP
(each = 90°)
APE = CPD
(vertically opposite angles)
 AEP ~ CDP
(AA similarity criterion)
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(ii) In ABD and CBE, we have
D
ADB = CEB
(each = 90°)
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
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ABD = CBE
(common)
 ABD ~ CBE
(AA similarity criterion)
(iii) In AEP and ADB, we have
A = A
(common)
AEP = ADB
(each = 90°)
 AEP ~ ADB
(AA similarity criterion)
(iv) In PDC and BEC, we have
PCD = BCE
(common)
CDP = CEB
(each = 90°)
 PDC ~ BEC
(AA similarity criterion)
MATHEMATICS–X
TRIANGLES
P
A
E
B
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Example 6. CD and GH (D and H lie on AB and FE) are respectively bisectors of ACB and EGF and
ABC ~ FEG. Prove that : (i)
Solution.
(i) Given ABC ~ FEG
 A = F
and C = G
...(1)


CD AC
(ii) DCB ~ HGE (iii) DCA ~ HGF.
=
GH FG
C
G
2 1
1
1
C  G
2
2
1= 3
and 2 =  4

AC CD
=
FG GH
B
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3 4
B F
A
D
H
...(2)
[ CD and GH are bisector of C and G respectively]
Now, In ACD and FGH, we have
A= F
[from (1)]
2 = 4
[from (2)]
 ACD ~ FGH (AA similarity criterion)
 DCA ~ HGF
(ii) We have, ACD ~ FGH
[NCERT]
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(iii) In DCB and HGE, we have
1 = 3
[from (2)]
B = E
( ABC ~ FEG)
 DCB ~ HGE
(AA Similarity criterion)
OA OD

Example 7. In the given figure,
. Prove that A = C and B = D
OC OB
Solution.
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In AOD and COB, we have
OA OD
=
OC OB
OA OC
=
OD OB
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
and, 1 = 2
( given)
( vertically opposite angles)
( SAS similarity criterion)

AOD ~ COB

A = C and B = D
Example 8. Two poles of height a metres and b metres are p metres apart. Prove that the height of the point
of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by
ab
metres.
ab
90
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MATHEMATICS–X
Solution.
Let AB and CD be two poles of height a metres and b metres respectively such that poles are p
metres apart. Let the lines AD and BC meet at O such that OE = h metres.
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D
B
b
O
a
h
C
x
E
y
A
Let CE = x and EA = y, so that x + y = p.
Now, In ABC and EOC, we have
CAB = CEO
(each = 90°)
C = C
(common)
 CAB ~ CEO
(AA similarity criterion)
CA AB
p a
ph

  x
...(1)
CE EO
x h
a
In AEO and ACD, we have
AEO = ACD
(each = 90°)
A = A
(common)
 AEO ~ ACD
(AA similarity criterion)


AE OE
y h
ph

   y
AC CD
p b
b
Now, x + y = p


B
ph ph
[using (1) and (2)]

p
a
b
1 1 1
1 1
ph     p   
h a b
a b
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...(2)
1 ab
ab

h
h
ab
ab
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the

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ab
metres.
ab
Example 9. Prove that area of an equilateral triangle described on one side of a square is equal to half the
area of the equilateral triangle described on one of its diagonal.
[NCERT]
Solution. Let ABCD be the given square. Equilateral triangles BCE and ACF are described on side BC
and diagonal AC respectively.
opposite pole is
F
D
C
E
A
B
Since BCE and ACF are equilateral they are equiangular and hence BCE ~ ACF
(AAA similarity criterion)
MATHEMATICS–X
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
ar ( BCE) BC 2

ar ( ACF) AC 2

BC2


[ ABCD is a square  diagonal =
( 2BC) 2
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2 (side)  AC =
2 . BC]
BC2
1

2
2.BC
2
ar (BCE) 1
 . Hence proved.
ar (ACF) 2
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Example 10. In the given figure, XY || AC and XY divides triangular region ABC into two parts equal in area.
AX
.
AB
We have, XY || AC and ar (BXY) = ar (quad. XYCA)
 ar (ABC) = 2.ar (BXY)
...(1)
Determine
Solution.
Now, XY || AC and BA is a transversal.
 BXY= BAC
...(2)
Thus, In BAC and BXY, we have
XBY = ABC
(common)
BXY = BAC
[from (2)]
 BAC ~ BXY
(AA similarity criterion)

ar ( BAC) BA 2

ar ( BXY) BX 2

2

BA  2.BX

BA  2(BA–AX)

( 2  1)BA  2AX
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BA 2
BX 2
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B
Y
C
2 –1
Ans.
2
Example 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same
time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per

AX
=
AB
B
[using (1)]
A
hour. How far apart will be the two planes after 1
Solution.
Distance travelled by an aeroplane due north in 1

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[NCERT]
1
3
hours  1000  km = 1500 km
2
2
OA = 1500 km
Distance travelled by an aeroplane due west in 1

1
hours?
2
1
3
hours = 1200 × km = 1800 km
2
2
OB = 1800 km
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MATHEMATICS–X
Now, In right angled AOB,
AB2 = OA2 + OB2
(using pythagoras theorem)
2
2
= (1500) + (1800)
= 2250000 + 3240000 = 5490000

AB  5490000 km  3  3  61 100  100 km
= 300
A
North
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1500 km
B
1800 km
West
61 km Ans.
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Example 12. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of
its diagonals.
[NCERT]
OR
2
ABCD is a rhombus. Prove that AB + BC2 + CD2 + DA2 = AC2 + BD2.
[CBSE 2005]
Solution. Let the diagonals AC and BD of rhombus ABCD intersect at O. Since the diagonals of a rhombus
bisect each other at right angle.

Since AOB is a right triangle right-angled at O.

AB2 = OA2 + OB2
2
2
B
...(1)
Similarly, we have
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4BC2 = AC2 + BD2
4CD2 = AC2 + BD2
2
A
C
O
B
[ OA = OC and OB = OD]
4AB2 = AC2 + BD2
2
D
(using Pythagoras theorem)
1
 1

  AC    BD 
2
2

 


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AOB = BOC = COD = DOA = 90° and AO = CO, BO = DO.
2
and 4AD = AC + BD
adding all these results, we get
...(2)
...(3)
...(4)
4 (AB2 + BC2 + CD2 + AD2) = 4 (AC2 + BD2)

AB2 + BC2 + CD2 + AD2 = AC2 + BD2
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Hence proved.
Example 13. In an equilateral triangle ABC, D is a point on side BC such that BD = 1 BC. Prove that
3
2
2
9AD = 7AB .
[NCERT]
Solution. Draw AP  BC.
Since AP  BC  BP 
1
1
BC and BD  BC
2
3
A
1 1
1
 DP     BC  BC
6
 2 3
Now, In right angled APB,
AB2 = BP2 + AP2

AB2 = BP2 + AD2 – DP2
MATHEMATICS–X
B
D
P
C
( AP2 = AD2 – DP2)
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2

1

1

AB2   BC   AD 2   BC 
2

6


AB2 
2
1
1
AB2  AD 2  AB2
4
36
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( AB = BC = AC)
9AB2 +36AD2 –AB2
36
2
2
 36AB = 8AB + 36AD2
 28AB2 = 36AD2
 7AB2 = 9AD2
Hence proved.
Example 14. In the given figure, O is a point in the interior of a triangle ABC, OD  BC, OE  AC and
OF  AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

AB2 =
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Solution.
(i)
In right OFA, ODB and OEC,
using Pythagoras theorem, we have
OA2 = AF2 + OF2
OB2 = BD2 + OD2
2
2
B
...(1)
...(2)
2
OC = CE + OE
...(3)
Adding (1), (2) and (3), we get
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A
F
B
[NCERT]
E
O
D
C
OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
which proves the (i) part.
(ii) In right ODB and ODC, we have
OB2 = OD2 + BD2
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and
OC2 = OD2 + CD2

OB2 – OC2 = BD2 – CD2
...(4)
Similarly, we have
OC2 – OA2 = CE2 – AE2
and
2
2
2
2
OA – OB = AF – BF
...(5)
...(6)
Adding (4), (5) and (6), we get
(OB2 – OC2)+ (OC2 – OA2) + (OA2 – OB2)= (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2)

0 = (BD2 + CE2 + AF2) – (AE2 + CD2 + BF2)

AF2 + BD2 + CE2 = AE2 + CD2 + BF2 , which proves the (ii) part.
Example 15. ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of
perpendicular from C on AB, prove that :
(i) cp = ab
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(ii)
1
1
1
= 2+ 2
2
p
a b
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[CBSE 2002]
MATHEMATICS–X
Solution.
(i) ar (ABC) 
1
 AB  CD
2
A
1
 cp
2
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...(1)
Dc
1
also, ar (ABC)   BC  AC
2
p
B
1
ab
2
from (1) and (2), we get

...(2)
1
1
cp  ab  cp  ab, which proves (i) part.
2
2
Since ABC is a right triangle right-angled at C.
 AB2 = BC2 + AC2  c2 = a2 + b2
2
 ab 
    a 2  b2
 p
a 2b 2

 a2  b2
p2

1
a2  b2

p2
a 2b 2
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a
b
C
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
ab 
 cp  ab  c 

p


1
a2
b2
 2 2 2 2
2
p
ab
ab

1
1 1
 
p2 b2 a2

1
1 1

 , which proves the (ii) part.
p 2 a 2 b2
PRACTICE EXERCISE
Questions based on Basic Proportionality Theorem :
1. In the given figure, DE || BC and
MATHEMATICS–X
AD 3
 . If AC = 4.8 cm, find AE.
DB 5
[CBSE 2002, 2003]
A
D
B
E
C
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2. In the given figure, find the value of x if DE || AB.
C
4
D
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x–3
E
3x–19
x–4
A
B
3. In the given figure, DE || BC. If :
A
(i) AD = 4 cm, DB = 6 cm and AC = 12.5 cm, find AE.
(ii) AD : DB = 3:5 and AE = 3.6 cm, find AC.
E
D
(iii) AE : EC = 5 : 3 and AB = 15.6 cm, find DB.
(iv) AE : AC= 5 : 9 and DB = 7.2 cm, find AB.
B
C
(v) If AB = 2 AD and AE = 4.8 cm, find EC.
4. In a ABC, D and E are points on the sides AB and AC respectively such that DE || BC.
(i) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x.
(ii) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.
[CBSE 2002]
(iii) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find x.
(iv) If BD = x – 3, AB = 2x, CE = x – 2, AC = 2x + 3, find x.
5. In a ABC, D and E are points on the sides AB and AC respectively. For each of the following cases
show that DE || BC :
(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
(ii) AB = 11.2 cm, DB = 2.8 cm, AC = 14.4 cm and AE = 10.8 cm
(iii) AB = 24 cm, AD = 15 cm, AE = 22.5 cm and AC = 36 cm
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm
6. In a ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm,
DE = 2 cm and BC = 5 cm, find BD and CE.
[CBSE 2001C]
7. M and N are points on the sides PQ and PR respectively of PQR. For each of the following cases, state
whether MN || QR.
(i) PM = 12 cm, QM = 15 cm, PN = 10 cm and NR = 12.5 cm
(ii) PQ = 1.8 cm, QM = 1.2 cm, PN = 2.7 cm and PR = 3.6 cm
(iii) PQ = 1.25 cm, PR = 2.5, PM = 0.17 cm and PN = 0.34 cm
(iv) PQ = 30 cm, PR = 36 cm, PM = 12 cm and NR = 21.6 cm.
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8.
In the given figure, EF || AB || DC. Prove that
B
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AE BF
=
.
ED FC
A
E
B
F
D
C
OR
Prove that a line-segment drawn parallel to the parallel sides of a trapezium divides the non-parallel sides
proportionally.
96
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MATHEMATICS–X
9. Let X be any point on the side BC of ABC (see figure given below). If XM, XN are drawn parallel to BA
and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB × TC.
A
T
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A
M
N
X
B
10. In the given figure, DE || AC and DC || AP. Prove that
BE BC

.
EC CP
A
D
C
[CBSE 2005]
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P
C
E
11. Let ABC be a triangle and D and E be two points on side AB such that AD = BE (see figure). If DP || BC
and EQ || AC, then prove that PQ || AB.
A
D
P
B
B
E
C
Q
Any point X inside DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE
meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF.
If three or more parallel lines are intersected by two transversals, prove that the intercepts made by them
on the transversals are proprotional.
ABC is a triangle in which AB = AC. Points D and E are points on the sides AB and AC respectively such
that AD = AE. Show that the points B, C, E and D are concyclic (lie on the same circle).
D
C
Use the given figure to find the value of x if :
B
12.
13.
14.
15.
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T
I
(i) OA = 4x – 2, OB = 4x + 1, OC = 2x + 2, OD = 3x – 1
(ii) OA = 3x – 1, OB = 2x + 1, OC = 5x – 3, OD = 6x – 5
O
B
A
16. In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.
P
A
O
B
C
R
Q
17. ABCD is a quadrilateral. P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C. Prove that PQRS is a parallelogram.
MATHEMATICS–X
TRIANGLES
97
18. In the given figure, D is a point on AB and E is a point on AC of ABC such that DE || BC. If AD : DB =
1:2, AB = 18 cm and AC = 30 cm, find the values of AE, EC, AD and DB.
A
J
A
E
D
C
B
19. In the given figure, DE || AB and FE || DB.
Prove that DC2 = CF . AC
J
A
[CBSE 2007]
C
F
E
D
B
A
20. In the given figure, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.
A
D
T
I
B
E
B
C
21. In the given figure, A = B and AD = BE. Show that AB || DE.
C
D
E
A
B
22. Using basic proportionality theorem, prove that line segment joining the mid-points of any two sides of
a triangle is parallel to the third side.
23. Determine a point on a line segment AB = 8 cm, which divides AB internally in the ratio 2 : 3. Prove your
assertion.
24. In the given figure, ABC is a triangle; and P, Q and R are points on BC, CA and AB respectively such that
PQ || AB and QR || BC. Prove that RP || CA, if R is the mid-point of AB.
A
M
A
R
Q
B
C
P
25. ABCD is a parallelogram. P is a point on side BC, and DP when produced meets AB produced at L. Prove
that :
(a) DP : PL = DC : BL
(b) DL : DP = AL : DC
98
TRIANGLES
MATHEMATICS–X
D
C
J
A
P
A
L
B
Questions based on Criteria for similarity of Triangles :
26. In the given figure, PQ || RS. Find PT and QT.
P
T
3 cm
cm
3.8
3.2
cm
Q
R
2 cm
S
J
A
27. In the given figure, ABC = 90° and BD  AC. If BD = 8cm and AD = 4 cm, find CD.
A
D
8c
m
4c
m
B
C
B
28. In the given figure, AB || QR. Find PB.
T
I
P
A
Q
6c
m
3 cm
9 cm
B
R
29. In the given figure, DE || BC, AD = 2 cm, BD = 2.5 cm, AE = 3.2 cm and DE = 4 cm. Determine AC and BC.
M
A
A
2 cm
D
2.5 cm
3.2 cm
4 cm
E
B
C
30. In the given figure, find x in terms of a, b and c.
A
E
a
53°
x
53°
C
c
b
D
31. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find
AB.
MATHEMATICS–X
B
TRIANGLES
99
32. A vertical stick 15 m long casts its shadow 10 m long on the ground. At the same time, the flag pole casts
a shadow 60 m long. Find the height of the pole.
33. In the given figure, AQ and PB are perpendiculars to AB. If AO = 10 cm, BO = 6cm and PB = 9 cm, find AQ.
J
A
P
A
B
O
Q
J
A
34. P and Q are points on the sides AB and AC respectively of a triangle ABC. If AP = 3 cm, PB = 6 cm,
1
AQ = 4.5 cm and QC = 9 cm; prove that PQ  BC.
3
35. In the given figure, ABC is a triangle, right angled at B. If FG || DE || CB and AG = GE = EB, find DE + FG.
C
D
F
B
12 cm
A
B
E
G
36. In the given figure, EDC ~ EBA, BEC = 115° and EDC = 70°. Find :
(i) DEC
D
C
70°
(ii) DCE
(iii) EAB
E 115°
(iv) AEB
(v) EBA
B
A
AO BO 1
37. In the given figure,
=
 and AB = 5 cm. Find DC.
OC OD 2
T
I
M
A
5 cm
A
B
O
D
C
38. In ABC, DE is parallel to base BC, with D on AB and E on AC. If
100
AD 2
BC
 , find
. [CBSE 2002C]
DB 3
DE
A
D
B
TRIANGLES
E
C
MATHEMATICS–X
39. In the given figure, AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm. Find x
and y.
F
B4 c
m
D
yc
m
x cm
6 cm
J
A
10 cm
A
C
E
40. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is
3.6 m above the ground, find the length of her shadow after 4 seconds.
J
A
CA CB
=
or, CA2 = CB × CD.
CD CA
[CBSE 2004]
42. Two right triangles ABC and DBC are drawn on the same hypotenuse BC on the same side of BC. If AC
and DB intersect at P, prove that AP × PC = BP × PD.
[CBSE 2000]
2
43. In a ABC, AB = AC and D is a point on side AC, such that BC = AC × CD. Prove that BD = BC.
41. D is a point on the side BC of ABC such that ADC = BAC. Prove that
44. In the given figure, BC  AB, DE  AC and DA  AB. Prove that : DE × BC = AE × AB.
C
D
B
E
B
A
45. In the given figure, PA, QB and RC are each perpendicular to AC and AP = x, QB = z, RC = y, AB = a and
BC = b. Prove that
T
I
1 1 1
  .
x y z
P
x
M
A
R
Q
z
y
a
A
B b C
46. In ABC, if AD  BC and AD = BD × CD, prove that BAC = 90°.
2
[CBSE 2004, 2006]
A
B
C
D
47. In a right angled triangle ABC, A = 90° and AD  BC. Prove that AD2 = BD × CD. [CBSE 2004 (C)]
48. In the given figure, AB || DE and BD || EF. Prove that DC2 = CF × AC.
[CBSE 2004 (C)]
C
F
E
D
MATHEMATICS–X
A
B
TRIANGLES
101
49. In an isosceles ABC, the base AB is produced both ways to P and Q such that AP × BQ = AC2. Prove
that ACP ~ BCQ.
J
A
C
P
Q
A
B
50. ABCD is a trapezium with AB || DC. If AED ~ BEC, prove that AD = BC.
D
C
E
J
A
B
A
51. In a quadrilateral ABCD, AB || CD and E and F are the mid-points of AB and CD respectively. If the
diagonal BD intersects EF in G, prove that DF.BG = EB.GD.
52. In a given figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at F, where E is any
point on the side BC. Prove that DF.EF = BF.AF.
B
D
A
F
T
I
B
C
E
QT QR
53. In the given figure, PR  QS and 1 = 2. Prove that PQS ~ TQR.
M
A
Q
T
P
1
2
R
S
54. CD and GH are respectively the medians of ABC and EFG. If ABC ~ FEG, prove that
(i) ADC ~ FHG
(ii)
CD AB
=
GH FE
(iii) CDB ~ GHE
55. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC
in L and AD produced in E. Prove that EL = 2.BL.
102
D
A
L
B
E
M
C
TRIANGLES
MATHEMATICS–X
56. Prove that the line segments joining the mid-points of the sides of a triangle form four triangle, each of
which is similar to the original triangle.
J
A
57. In the given figure, FEC  GDB and 1 = 2. Prove that ADE ~ ABC.
A
1 2
D
E
J
A
G
C
B
58. If the angles of one triangles are respectively equal to the angles of another triangle, prove that the ratio
of their corresponding sides is the same as the ratio of their corresponding :
(i) medians
(ii) angle bisectors
(iii) altitudes
59. In the given figure, ABC is a right triangle right angled at B and D is the foot of the perpendicular drawn
from B on AC. If DM  BC and DN  AB, prove that :
F
(i) DM2 = DN × MC
(ii) DN2 = DM × AN
A
D
N
T
I
B
C
B
M
60. In the given figure, DEFG is a square and BAC = 90°. Prove that :
(i) AGF ~ DBG
(ii) AGF ~ EFC
(iii) DBG ~ EFC
(iv) DE2 = BD × EC
A
M
A
G
B
D
F
E
C
Questions based on Areas of similar Triangles :
61. D and E are points on the sides AB and AC respectively of ABC such that DE is parallel to BC and
AD : DB = 4:5. CD and BE intersect each other at F. Find the ratio of the areas of DEF and BCF.
[CBSE 2003]
2
2
62. The areas of two similar triangles are 100 cm and 49 cm respectively. If the altitude of the bigger triangle
is 5 cm, find the corresponding altitude of the other.
[CBSE 2002]
2
2
63. The areas of two similar triangles are 121 cm and 64 cm respectively. If the median of the first triangle is
12.1 cm, find the corresponding median of the other.
[CBSE 2001]
64. In ABC, P divides the side AB such that AP : PB = 1 : 3. Q is a point on AC such that PQ || BC. Find the
ratio of the areas of APQ to trapezium BPQC.
MATHEMATICS–X
TRIANGLES
103
65. ABC is an isosceles triangle right angled at B. Two equilateral triangles are constructed with side BC and
1
AC as shown. Prove that : ar (BCD)  ar (ACE) .
[CBSE 2001]
2
E
C
J
A
D
A
B
ar (DFE)
66. In the given figure, DE || BC and AD : DB = 5 : 4. Find ar (CFB) .
A
D
E
F
B
J
A
[CBSE 2000]
C
67. In the given figure, DE || BC and DE : BC = 4 : 5. Find ratio of the areas of ADE and trapezium BCED.
A
B
E
D
C
B
T
I
68. If the area of two similar triangles are equal, prove that they are congruent.
69. Prove that area of an equilateral triangle described on the side of a square is half the area of the
equilateral triangle described on its diagonal.
70. Equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the
hypotenuse is equal to the sum of the areas of triangles on the other two sides.
[CBSE 2002]
71. ABCD is a trapezium in which AB || DC and AB = 2DC. Determine the ratio of the areas of AOB and
COD.
72. In thetrapeziumABCD, AB || CD andAB = 2CD. If thearea of AOB = 84 cm2, find the area of COD.
[CBSE 2005]
73. D and E are points on the sides AB and AC respectivley of a ABC such that DE || BC and divides ABC
BD
.
into two parts, equal in area, find
[CBSE 2000]
AB
74. In the given figure, ABC and DBC are on the same base BC. If AD and BC intersect at O, prove that
M
A
ar (ABC) AO

.
ar (DBC) DO
104
[CBSE 2002, 2005]
C
A
O
B
TRIANGLES
D
MATHEMATICS–X
75. In ABC, D is a point on AB, E is a point on AC and DE || BC. If AD : DB = 5 : 4, find :
A
(i) DE, if BC = 27 cm
D
(ii) ar (ADE) : ar (ABC)
C
B
(iii) ar (ADE) : ar (trapezium DBCE)
J
A
E
Questions based on Pythagoras Theorem and its converse
76. The sides of a some triangles are given below. Determine which of them are right triangles.
(i) 15 cm, 20 cm, 25 cm
(ii) 5 cm, 12 cm, 13 cm
(iii) 7 cm, 24 cm, 25 cm
(iv) 5 cm, 8 cm, 12 cm
(v) 8 cm, 15 cm, 17 cm
(vi) 9 cm, 40 cm, 41 cm
77. Show that the ABC is right triangle, if :
(i) AB = 24 cm, BC = 25 cm and AC = 7 cm
(ii) AB = 12 cm, BC = 16 cm and AC = 20 cm
78. In PQR, Q = 90°, find :
(i) PQ, if PR = 34 cm and QR = 30 cm
(ii) PR, if PQ = 24 cm and QR = 7 cm
J
A
79. In PQR, QPR = 90° and QR = 26 cm. If PS  SR, PS = 6 cm and SR = 8 cm, find the area of PQR.
P
6c
m
B
S8
Q
cm
R
26 cm
80. A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the
foot of the ladder from the building.
81. A ladder 25 m long reaches a window which is 24 m above the ground on side of the street. Keeping the
foot at the same point, the ladder is turned to the other side of the street to reach a window 7 m high. Find
the width of the street.
82. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance betwen their feet is 12 m, find
the distance between their tops.
[CBSE 2002 C]
T
I
83. In an equilateral ABC, AD  BC, prove that AD2 = 3BD2.
84. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.
85. In an equilateral triangle with side a, prove that :
M
A
3
a
(i) Altitude 
2
[CBSE 2001 C]
3 2
a
[CBSE 2002]
4
86. In the given figure, ABC is a right triangle right-angled at B. AD and CE are the two medians drawn from
(ii) Area 
3 5
cm. Find the length of CE.
A and C respectively. If AC = 5 cm and AD 
2
A
MATHEMATICS–X
E
B
D
C
TRIANGLES
105
87. ABC is an isosceles triangle right angled at C. prove that AB2 = 2AC2.
[CBSE 2005]
88. ABC is an isosceles triangle with AB = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
[CBSE 2004 (C)]
2
2
2
2
89. In a quadrilateral ABCD, CA = CD, B = 90° and AD = AB + BC + CA . Prove that ACD = 90°.
D
J
A
C
J
A
A
B
90. In the given figure, BCA = 90°. Q is the mid-point of BC. Prove that : AB2 = 4AQ2 – 3AC2.
[CBSE 2004 (C)]
A
C
B
Q
91. In the given figure, ABCD is a square. F is the mid-point of AB, BE is one-third of BC. If the area of FBE
= 108 cm2, find the length of AC.
D
C
T
I
B
E
B
A
F
92. Prove that three times the square of any side of an equilateral triangle is equal to four times the square
of the altitude.
[CBSE 2002]
2
2
2
93. ABC is an obtuse triangle, obtuse angled at B. If AD  CB, prove that AC = AB + BC + 2BC × BD.
94. B of ABC is an acute angle and AD  BC, prove that AC2 = AB2 + BC2 – 2BC × BD.
95. P and Q are the mid-points of the sides CA and CB respectively of a ABC, right angled at C. Prove that:
(i) 4 AQ2 = 4AC2 + BC2
(ii) 4BP2 = 4BC2 +AC2
[CBSE 2001]
(iii) 4 (AQ2 + BP2) = 5AB2
[CBSE 2001, 2006 (C)]
96. ABC is a triangle in which AB = AC and D is any point in BC. Prove that AB2 – AD2 = BD . CD.
[CBSE 2005]
97. The perpendicular AD on the base BC of a ABC intersects BC at D so that DB = 3CD. Prove that
2AB2 = 2AC2 + BC2.
[CBSE 2005]
98. ABC is a right triangle right-angled at B. Let D and E be any points on AB and BC respectively. Prove that
AE2 + CD2 = AC2 + DE2.
[CBSE 2002 (C)]
M
A
99. In a ABC, AD  BC and AD2 = BC × CD. Prove that ABC is a right triangle. [CBSE 2006 (C), 2007]
100. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of
its sides.
101. A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that
OB2 + OD2 = OC2 + OA2.
[CBSE 2006 (C)]
106
TRIANGLES
MATHEMATICS–X
102. In the given figure, D and E trisect BC. Prove that :
8AE2 = 3AC2 + 5AD2
A
[CBSE 2006(C)]
J
A
C
D
B
E
2
2
2
2
103. ABCD is a rhombus. Prove that : AB + BC + CD + DA = AC2 + BD2.
[CBSE 2005, 2006]

104. In a ABC, AC > AB, D is the mid-point of BC and AE BC. Prove that :
A
1
(i) AC2 = AD2 + BC.DE + BC2
4
1
(ii) AB2 = AD2 – BC.DE + BC2
4
B
C
E D
1
(iii) AB2 + AC2 = 2AD2 + BC2
[CBSE 2006(C)]
2
105. P and Q are points on the side CA and CB respectively of ABC, right angled at C, prove that :
AQ2 + BP2 = AB2 + PQ2.
B
J
A
HINTS TO SELECTED QUESTIONS
9. In TXM, XM || BN 
TB TN

TX TM
In TMC, XN || CM 
TX TN

TC TM
...(1)
T
I
from (1) and (2), we will get desired result.
17. In ADC,
DS 2SA
DR 2RC

 2 and,

2
SA SA
RC
RC
M
A

...(2)
DS DR

 SR || AC .
SA RC
D
R
C
Q
S
A
Similarly, PQ || AC
Also, by joining BD, we can prove that PS || RQ.
25. (a) In ALD, BP || AD 
B
BL LP

. Now use the fact that AB = CD.
BA PD
41. In ABC and DAC,
ADC = BAC and C= C  ABC ~ DAC

P
( AA similarity)
AB BC AC
=
=
etc.
DA AC DC
45. Clearly, ABQ ~ ACR 
MATHEMATICS–X
z
a

y a b
...(1)
TRIANGLES
107
also, CBQ ~ CAP 
z
b

x ab
...(2)
J
A
adding (1) and (2), we get
1 1
z z a b
1 1 1
 
 z   1  
y x a b
x
y
x y z


( AAS congruence rule)
55. Clearly, BMC  EMD

BC = DE
( cpct)
Then, AE = 2BC

EL AE 2BC


 2  EL  2BL
BL CB BC
59. Clearly, quadrilateral BMDN is a rectangle. Then, BM = ND
here, BMD ~ DMC
( AA similarity)
Also, BND ~ DNA
( AA similarity)
74. Draw AL  BC and DM  BC.


AL AO

DM DO
...(1)
T
I
1
 BC  AL
ar (ABC) 2
AL AO



ar (DBC) 1
DM DO
 BC  DM
2
75. (i) In ABC, DE || BC  ADE ~ ABC.
2
(ii)
[ using (1)]
AD DE
5
DE



 DE  15cm
AB BC
5  4 27
M
A

B
( AA similarity)
Clearly, ALO ~ DMO
2
ar (ADE) AD
5

    25 : 81
2
ar (ABC) AB
9
(iii) Consider,

108
J
A
( AA Similarity)
Also, AEL ~ CBL
ar (trap. DBCE) ar (ABC)  ar(ADE)

ar (ADE)
ar (ADE)

ar (ABC)
81
56
1 
1 
ar (ADE)
25
25
[ using (ii)]
ar (ADE)
 25 : 56
ar (trap. DBCE)
TRIANGLES
MATHEMATICS–X
93. In ADB, AB2 = AD2 + DB2
A
...(1)
In ADC, AC2 = AD2 + DC2
J
A
= AD2 + (DB + BC)2
= (AD2 + DB2) + BC2 + 2BC.BD
[ using (1)]
= AB2 + BC2 + 2BC.BD
D
C
B
1
2
100. We know that if AD is a median of ABC, then AB2 + AC2 = 2AD2  BC .
2
Since diagonals of a parallelogram bisect each other.
 BO and DO are medians of ABC and ADC respectively.

J
A
O
1
AC 2
2
AB2 + BC2 = 2 BO2 +
D
...(1)
A
1
AC 2
2
and, AD2 + CD2 = 2DO2 +
...(2)
add (1) and (2), and simplify now.
B
102. Here, BD = DE = CE = x (say). Then, BE = 2x and BC = 3x.
In right ABD, ABE and ABC,
T
I
AD2 = AB2 + x2
...(1)
AE2 = AB2 + 4x2
and
2
2
...(2)
2
AC = AB + 9x
...(3)
C
B
Consider, 8 AE2 – 3 AC2 – 5AD2. Using (1), (2) and (3) in this expression, we get
8AE2 – 3AC2 – 5AD2 = 0  8AE2 = 3AC2 + 5AD2.
M
A
104. We have AED = 90°, ADE < 90° and ADC > 90°.
(i) In ADC, ADC is an obtuse angle.
 AC2 = AD2 + DC2 + 2DC.DE
2
1
1

 AD 2   BC   2. BC.DE. Simplify now..
2
2

(ii) In ABD, ADE is an acute angle.

AB2 = AD2 + BD2 – 2BD . DE
2
1
1

 AD 2   BC   2. BC.DE. Simplify now..
2
2

(iii) Adding (i) and (ii), we will get the desired result.
MATHEMATICS–X
TRIANGLES
109
MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
J
A
1. D and E are respectively the points on the sides AB and AC of a ABC and DE || BC, such that AD = 3
cm, AB = 12 cm, AE = 4 cm, then the value of CE is :
(a) 6 cm
(b) 9 cm
(c) 12 cm
(d) 15 cm
2. In the given figure, PQ || DC || AB. The value of x is :
A
x
B
x–1
Q
P
x+2
C
D
(a) 1
(b) 2
(c) 3
3. In the given figure, DE || BC, BE || XC and
(d) 4
AD 2
AX
 , then the value of
is:
DB 1
XB
A
B
E
D
B
C
X
(a) 2 : 1
(b) 3 : 1
(c) 3 : 2
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I
4. In the given figure, AB || CD. Then the value of x is :
D
4
3x
–
C
3
x–
B
(b) 11
M
A
(d) none of these
x+
4
19
A
(a) 8
J
A
x
(c) 8 or 11
(d) none of these
5. In the given figure, ACB ~ APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, the value of AC
is:
B
P
(a) 5.6 cm
A
Q
C
(b) 2.8 cm
(c) 8 cm
(d) none of these
6. A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the
shadow 40 m long on the ground. The height of the tower is :
(a) 50 m
(b) 60 m
(c) 80 m
(d) none of these
7. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 8 cm, then
the value of AB is :
(a) 12 cm
110
(b) 15 cm
(c) 18 cm
TRIANGLES
(d) none of these
MATHEMATICS–X
J
A
5
3 cm
cm
8. In the given figure, BAC = ADB= 90°. The length of DC is :
A
B 4 cm
C
D
(a) 2 cm
(b) 2.25 cm
(c) 2.5 cm
(d) 2.75 cm
9. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, the length of QR is:
(a) 5 cm
(b) 5.5 cm
(c) 6 cm
(d) none of these
10. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle
is 4.5 cm, the length of corresponding altitude of smaller triangle is :
(a) 3 cm
(b) 4 cm
(c) 3.5 cm
(d) 4.5 cm
11. Two isosceles triangles have equal vertex angles and their areas are in the ratio 9 : 16. The ratio of their
corresponding altitudes is :
(a) 3 : 4
(b) 9 : 16
(c) 81 : 256
(d) none of these
J
A
12. In the given figure, DE || BC and AD : DB = 4 : 5. Then the value of
A
F
B
14.
15.
C
(a) 4 : 5
(b) 4 : 9
(c) 16 : 81
(d) none of these
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas
of the triangles ABC and BDE is :
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
D, E and F are the mid-points of the sides BC, CA and AB respectively of a ABC. The ratio of the areas
of DEF and ABC is :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Two vertical poles are 24 m apart. If the heights of the poles be 17 m and 27 m; the distance between the
tips of the poles is :
(a) 20 m
(b) 26 m
(c) 28 m
(d) none of these
Each side of a rhombus is 13 cm. If one of the diagonals is 24 cm, the length of the other diagonal is :
(a) 5 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
The area of a right-angled triangle is 60 cm2. If the larger side of the triangle exceeds the smaller side by
7 cm, the length of hypotenuse is :
(a) 10 cm
(b) 12 cm
(c) 15 cm
(d) 17 cm
In the given figure, the value of AD is :
D
T
I
M
A
16.
17.
18.
(a) 25 cm
MATHEMATICS–X
20 c
m
C
7 cm
13.
B
E
D
ar (DEF)
is :
ar (BCF)
B
(b) 15 cm
A
24 cm
(c) 30 cm
TRIANGLES
(d) none of these
111
19. In an equilateral triangle ABC, if AD  BC, then:
(a) 2AB2 = 3AD2
(b) 4AB2 = 3AD2
(c) 3AB2 = 4AD2
(d) 3AB2 = 2AD2
1
20. The perpendicular AD on the base BC of a ABC intersects BC at D so that CD  . DB.
4
(a) 3BC2 = 5 (AC2 – AB2)
(b) 3BC2 = 5 (AB2 – AC2)
(c) 5BC2 = 3 (AC2 – AB2)
(d) 5BC2 = 3 (AB2 – AC2)
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A
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. In figure, DE || BC. If
AD 4
AE
 , then what is the value of
?
DB 3
AC
A
E
D
B
C
2. In figure, XMN ~ XYZ, what is the measure of XZY?
X
M
75°
65°
B
N
Z
Y
T
I
3. In figure, DE || BC, what is the value of x?
M
A
A
3 cm
D
5 cm
B
J
A
5 cm
E
x
C
4. If the ratio of the corresponding sides of two similar triangles is 3 : 5, then what is the ratio of their
corresponding heights?
5. If the ratio of the corresponding medians of two similar triangles is 7 : 5, then what is the ratio of their
corresponding sides?
6. State the basic theorem of proportionality.
7. State the converse of the basic theorem of proportionality.
8. In figure, state whether EF || QR ?
P
112
3.5 cm
E
5 cm
4 cm
F
Q
TRIANGLES
6 cm
R
MATHEMATICS–X
9. In figure, AD is bisector of BAC, what is the value of DC.
A
J
A
5 cm
3 cm
B 10 cm
C
D
10. The perimeters of two similar triangles ABC and PQR are respectively 30 cm and 24 cm. If PQ = 8 cm, find
AB.
11. In figure, express x in the terms of a, b and c.
A
a
E
x
36°
b
B
D
36°
c
C
J
A
12. The lengths of sides of a triangle are 12 cm, 16 cm and 21 cm. The bisector of the greatest angle divides
the opposite side into two parts. Find the length of these two parts.
13. In figure, find AD.
C
3 cm
D
A
4 cm
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I
14. In figure, find x.
M
A
A
12 cm 30°
B
B
2 cm
x cm E
D
B
120°
9 cm
6 cm
C
15. The area of the two similar triangles are in the ratio 81 : 121. What is the ratio of their corresponding
sides?
16. If ABC ~ DEF such that BC = 5 cm, EF = 3 cm and ar (DEF) = 36 cm2, find the area of ABC.
17. If ABC ~ PQR such that ar (ABC) = 81 cm2 and ar (PQR) = 121 cm2. If QR = 22 cm, find BC.
18. If ABC ~ XYZ and the ratio of area of ABC to that of area of XYZ is 25 : 49, then what is the ratio
of their corresponding medians?
19. If ABC ~ PQR and their corresponding altitudes are in the ratio 7 : 8, then what is the ratio of the area
of PQR to that of area of ABC?
MATHEMATICS–X
TRIANGLES
113
20. State Pythagoras theorem.
21. State converse of Pythagoras theorem.
J
A
22. Is the triangle with sides 9 cm, 41 cm, 40 cm a right angled triangle?
23. Two poles of height 6 m and 11 m are standing 12 m apart. What is the distance between their tops?
24. In figure, ABC and DEC are the right triangles with B = E = 90°, find BE.
A
17
cm
8 cm
D
10
cm
E
C
B
25. In figure, ADB = BAC = 90°. What is the value of x?
A
6 cm
D
8 cm
B
26 cm
T
I
x
B
PRACTICE TEST
M.M : 30
General Instructions :
J
A
6 cm
C
Time : 1 hour
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. Determine, by using the pythagoras theorem, whether the triangles having sides (2a – 1) cm, 2 2a cm
and (2a + 1) cm is a right angled triangle.
M
A
2. Prove that the diagonals of a trapezium divide each other proportionally.
3. If ABC ~ DEF such that BC = 3 cm, EF = 4 cm and area of ABC= 54 cm2. Determine the area of DEF.
4. The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point
on the side BC. Prove that DF × FE = BF × FA.
5. In the given figure, PQ = PR, PN  QR and LM  PR. Prove that
114
PQ QN

.
LR MR
P
M
L
Q
TRIANGLES
N
R
MATHEMATICS–X
6. ABC is a right angled triangle in which C = 90°. If p is the length of the perpendicular from C to AB and
1
1
1
AB = c, BC = a, CA = b, prove that p 2  a 2  b2 .
J
A
7. In ABC, A = 90°, if AD  BC, prove that AB2 + CD2 = BD2 + AC2
8. D is a point on the side BC of ABC such that ADC = BAC. Prove that
CA CB
=
.
CD CA
1
9. In an equilateral ABC, D is a point on BC such that BD = BC. Prove that 9AD2 = 7AB2.
3
10. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the
other two sides are divided in the same ratio. Prove it.
J
A
Using above do the following : In the given figure, DE || BC and BD = CE. Prove that ABC is an isosceles
triangle.
A
E
D
B
B
C
ANSWERS OF PRACTICE EXERCISE
1. 1.8 cm
2. x = 8 or 11
3. (i) 5 cm
(ii) 9.5 cm (iii) 5.85 cm
T
I
(iv) 16.2 cm
4. (i) x = 11 (ii) x = 1 (iii) x = 4 (iv) x = 9
6. BD = 3.6 cm, CE = 4.8 cm
7. (i) yes (ii) no (iii) yes (iv) yes
15. (i) x = 5 (ii) x = 2 or x 
18. AE = 10 cm, EC = 20 cm, AD = 6 cm and DB = 12 cm
26. PT = 4.8 cm, QT = 5.7 cm
M
A
29. AC = 7.2 cm, BC = 9 cm
33. AQ = 15 cm
37. 10 cm
40. 1.6 m
66. 25 : 81
73.
2 2
2
61. 16 : 81
67. 16 : 9
27. CD = 16 cm
30. x 
ac
bc
35. 12 cm
38.
5
2
62. 3.5 cm
31. AB = 15 cm
32. 72 m
36. (i) 65 (ii) 45° (iii) 45° (iv) 65° (v) 70°
39. x = 3.75 cm, y = 6.67 cm
64. 1 : 3
2
71. 4 : 1
72. 21 cm
76. (i), (ii), (iii), (v), (vi)
78. (i) 16 cm (ii) 25 cm
79. 120 cm2
80. 15 m
82. 13 m
86. 2 5 cm
91. 12 2 cm
MATHEMATICS–X
1
2
28. PB = 2 cm
63. 8.8 cm
75. (i) 15 cm (ii) 25 : 81 (iii) 25 : 56
84. 13 cm
(v) 4.8 cm
TRIANGLES
81. 31 m
115
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (c)
2. (b)
3. (b)
4. (c)
5. (a)
6. (b)
7. (a)
8. (b)
9. (c)
10. (c)
11. (a)
12. (c)
13. (c)
14. (c)
15. (b)
16. (c)
17. (d)
18. (b)
19. (c)
20. (b)
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A
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
4
7
2. 40°
3. 25 cm
3
4. 3 : 5
9. 6 cm
10. 10 cm
11.
13. 1.5 cm
14. 8 cm
15. 9 : 11
16. 100 cm2
18. 5 : 7
19. 64 : 49
22. yes
23. 13 m
1.
8. No
25. 24 cm
ac
bc
B
J
A
5. 7 : 5
12. 9 cm, 12 cm
17. 18 cm
24. 23 cm
ANSWERS OF PRACTICE TEST
3. 96 cm2
1. yes
T
I
M
A
116
TRIANGLES
MATHEMATICS–X