1

Topic 1 - Congruency of
Triangles
Objective Type Questions On seeing this, certain questiong arises into
Meenal's mind. She drew the shape of cardboard
MCQs (1 Mark) poster on paper and marked different points and
1, In triangles ABC and DEF, AB = FD and angles. She observed that line segment joining
ZA=ZD. The two triangles will be congruent by the midpoint M and N of parallel sides AB and
SAS axiom, if CD respectively.
(a) BC= EF (b) AC= DE
(c) AC= EF (d) BC= DE (i) Length of AM is equal to length of
(a) MN (b) BM (c) AN (d) BN
(NCERT Exemplar)
(ii) By which congruency criterion, AAMN is
Fill in the Blanks (1 Mark) Congruent toABMN?
(a) SAS (b) RHS (c) SSS (d) ASA
2. In the given figure, AAOB = ACOB by
Congruence. (iii)21=
(a) Z3 (b) 24 (c) Z2 (d) D
(iv) ZAND =
(a) ZNAB (b) ZNMA
5 cm
B (c) ZNMB (d) ZBNC
(v) By which congruency criterion, AADN is
A congruent to ABCN?
(a) RHS (b) SSS (c) SAS (d) ASA
VSA Type Questions (1 Mark)
3. In the given figure, AOAP AOBP. State the Short Answer Type Questions
criteria by which the triangles are congruent. SA Type IQuestions (2 Marks)
6. In figure shown, two lines AB and CI
intersect each other at the point 0 such thai
BC | DA and BC = DA. Show that O is the mid
B
point of both the line-segments AB and CD.
4. In triangles ABC and DEF, ZA = ZD, ZB
= ZE and AB = EF. Will the two triangles be
congruent ? Give reasons for your answer.
(NCERT Exemplar)
D
Case StudyBased Questions (5x1 Mark) (NCERT Exemplar)
5. After reopening of school, students of class IX SA Tvpe Il Questions (3 Marks)
decided to make poster on corona virus pandemic
to spread awareness among school students. They 7. In the given figure, if AB=BC and ZA = 2C.
take a cardboard piece in the shape of trapezium Then find the value of x.
and divide it into three triangular parts and joined
it with a thin stick and wrote three slogans on it.
A M B 55°

Banitize E* 75°\D
Always yqur
hahds Keep
wear
HOcial
magk
distance A
D
8. In the given figure, if ADC = LAEC and
AB = BC, then prove that AE = CD.
 2
D A
B D

Long Answer Type Questions

LA Type Buestions (4 Marks) 10. Line segment joining the mid-points M and
Q. In the given figure, AB = AD, Z1 = /2 and N of parallel sides ABand DC, respectively of
3= 24. Prove that AP= AQ. a trapezium ABCDis perpendicular to both the
sides AB and DC. Prove that AD = BC.
(NCERT Exemplar)

SOLUTIONS
1, (b): In AABC and ADEF. (iv) (d) :Since, Z1= /2 ’ 90°- 1= 90°-Z2
AB =FD and ZA =D ZAND = LBNC ..(i)
A D (v) (c): In AADN and ABCN
ZAND = LBNC [From (ii)I
AN = BN [:: AAMN ABMN]
DN = CN [:Nis midpoint of CD]
AADN ABCN [By SAS congruency criterion]
We know that two triangles will be congruent by SAS 6. BC||AD [Given]
axiom, if two sides and the included angle of one triangle Therefore, ZCBO = DAO [Alternate interior angles]
are equal to the twosides and the included angle of other and ZBCO = ZADO [Alternate interior angles]
triangle. Also, BC= DA [Given]
AABC ADFE if AC= DE So, ABOC AAOD By (i), (ii), (ii) and ASA congruence]
Therefore, OB = OA and OC= 0D [By C.P.C.TJ
2. In AAOB and ACOB,
[Each equals 90] ’ O is the mid-point of both AB and CD.
ZBAO = LBCO
7. In AABD and ACBE,
LAOB = LCOB [Given]
OB = OB [Common] AB = B [Given
LBAD = BCE [Given]
AAOB ACOB [By AAS congruence] ZABD = ZCBE [Common]
3. In AOAP and AOBP, AABD ACBE [By ASA congruence]
OA = OB [Given] ZADB = ZCEB (By C.P.CI]
OP = OP [Common] But given ZADB=75° and ZCEB =y°
LAOP = ZBOP [Given] X=75
AOAP AOBP (By SAS congruence] ZBDC+ ZCDA= 180°
8. [Linear pair]
4
Iwo triangles need not be congruent, because AB ZBDC +x =180° ’ ZBDC = 180 -x
dnd EF are not corresponding sides in the two triangles. Similarly, ZBEA = 180° - y
D ) (b) : AM = BM I: M is the mid-point of AB] Since, x=y (Given]
...1) ZBDC = ZBEA ..)
In ABAE and ABCD
(1) (a):In AAMN and ABMN ZBEA = ZBDC
AM = BM (From ()] [From (1)]
ZB = ZB
23= 24
[90° each] |Common]
AB = BC
MN = MN [Common] (Given]
ABAE ABCD [By AAS congruence]
AAMN ABMN (By SAS congruency criterion] AE = CD
[By C.P.CT]
(i) (c):Since, AAMN ABMN Given AB= AD, 21= /2 and /3 = /4
.
[By C.P.C.T.] 9.
21= 2
3
21+ 23 = 2+ 24 A M
ZBAP + ZPAC = ZDAQ + ZQAC
ZBAC = 2DAC ...)
In ABAC and ADAC,
AB = AD
LBAC= LDAC
[Given]
[From (i)]
AC = AC
[Common D
ABAC ADAC (By SAS congruence] Now, in AAMN and ABMN,
ZABC = LADC
[By C.P.C.TJ ...(i) AM = BM [.:Mis the mid-point of AB]
Now, in AABP and AADQ, 23= Z4
AB = AD [Each equals 90°]
[Given] MN= MN
ZBAP = ZDAO
[::21=22(Given)] So, AAMN ABMN
[Common)]
ABP = ZADQ
[By SAS congruence]
[From (ü)] 21= 22 (By C.P.CI]
AABP = AADQ [By ASA congruence] 90°- Z1 =90°- 22 ’ ZAND =ZBNC
AP= AQ [By C.P.C.T] Now, in AADN and ABCN,
10. We have given a trapezium ABCD, points M ZAND = BNC [From ())
AN = BN [: AAMN ABMN
and N are the mid-points of parallel sides AB and CD DN= CN [:: Nis the mid-point of CD]
respectively. AADN ABCN [By SAS congruence]
Join MN, which is perpendicular to both AB and CD. Hence, AD = BC [By C.P.C.I]
4
Topic 2 - Properties of
Triangle

A
Objective Type Questions
MCQs (1 Mark)
1. In ABC, if BC=AB and ZA =80°, then C 125°
is equal to B
(a) 60 (b) 40° (c) 80° (d) 100°
7. In the given A
2. In the given figure, if PQ=QR, figure, AB = AC and
ZQPR = 48°, then PQR = 480 DB = DC. Prove that
(a) 48°
(b) 84° ZABD= LACD. B

(c) 30° 8. In an isosceles triangle ABC with AB = AC,

(d) 36° D and E are points on BC such that BE = CD
(see figure). Show that AD=AE.
Fillin the Blanks (1 Mark] A
3. Angles opposite to equal sides of a triangle
are

4. Sides opposite to angles of a triangle

are equal.
B (NCERT)
VSA Type Questions (1 Mark) E

5. In an isosceles triangle, if the vertex angle SA Type IlQuestions (3 Marks)

is twice the sum of the base angles, then find the 9. P is a point on the bisector of ZABC. If the
measure of each vertex angle of the triangle. line through P, parallel to BA meet BC at
prove that BPQ is an isosceles triangle.
Short Answer Type Questions (NCERT Exemplar)
SA Type l Questions (2 Marks) 10. ABCis a right triangle with AB=AC. Bisector
6. In the given figure, AB = AC and ACD of ZA meets BC at D. Prove that BC= 2AD.
= 125°. Find ZA. (NCERT Exenplar)
 5

11 ABC is an isosceles triangle in whieh Long Answer Type Ouestions

AC=BC.ADand BE are respectively two altitudes
to sides BC and AC. Prove that AE= B) LA Type Questions (4 Marks)
(NCERT Evemplur) 13. O is apoint in the interior of a square ABCD
12, In the given figur, if NT= NZ, ZYXA= LZXB. such that OAB is an equilateral triangle. Show
then provethat A\AB is an isosceles triangle. that AOCD is an isosceles triangle.
(NCERTEremplat)
14. Bisectors of the angles Band Cof an is09celes
triangle with AB = AC intersect each other
at 0. BO is produced to a point M. Prove that
A B ZMOC= LABC. (NCERT Exemplar)

SOLUTIONS
1. (c): We have, in AABC, BC= AB
ZABC - DBC = LACB- ZDCB
LC= LA [Angles opposite toequal sides of a ZABD= LACD
2C= 80°
triangle are equal] 8 In AABDand AACE,
[: ZA=80° (Given)] AB = AC
2 (b) : In APQR, PQ= QR [Given] .(i)
ZB= C
LQPR = 2QRP [Angles opposite to equal [Angles opposite to equal sides of a triangle
are equal] ..(ü)
sides of a triangle are equal] Also, BE = CD
LQRP = 48° : 2QPR= 48° (Given)] So, BE - DE = CD - DE’ BD = CE
Also,P+ 0+ ZQRP =180° [Angle sum property ...(11)
So, AABD AACE (By (i), (iü), (iii) and SAS congruence]
of a triangle] AD= AE
48° + Z0+48° =180°’ZQ= 180°- 96 ’ LQ=84° [By C.P.C.T]
9. We have given, P is
3. Angles opposite to equal sides of a triarngle are
equal. a point on the bisector of
4 Sides opposite to
ZABC and a line passes
equal angles of a triangle are through Pparallel to BA,
equal. meet BCat O.
5. Let each base angle of an isosceles triangle be x. We have, 1= 2
Given, vertical angle = 2(x+ x) = 4x [:: BP is bisector of ZB B
But x+ x + 4x= 180° [Angle sum property of a triangle] (Given)] C
6x= 180° ’x=30° As PQ|| ABand PB is a transversal.
Vertical angle =4(30°) =120° 1=23
6 So, 2= 23 [Alternate interior angles]
We have, ZACD=125°
ZACB = 180° - LACD ’ PQ= BQ [Sides opposite to equal angles of a
[Linear pair)
Thus, ABPQ is an isosceles triangle. triangle are equal]
ZACB= 180° - 125° = 55°
Also given,AB = AC
ZB= ZACB =55° [Angles opposite to equal 10. We have given, AABC which is an B
sides of a triangle are equal] isosceles right triangle with AB = AC
Now, ZA+ ZB + ZACB= 180° and AD is the bisector of ZA = 90°.
.". ZB = ZC [Angles
equal sides of a triangle areopposite
[Angle sum property of a triangle] to
ZA+ 55° + 55° = 180°
LA = 180°-110° = 70° Also, ZA+ ZB+ ZC=180° equal]
7. In ABC, we have [Angle sum property of a triangle]
AB = AC 90° +2ZB = 180° A
[Given] 2ZB =90° ’ ZB = 45° .: 23= 24= 45°
LABC = LACB ...) and 1= 12= 45° [* ADis bisector of ZAJ
IAngles opposite to equal sides of a triangle are equal] Also, Z1= 23= 22=24 = 45°
Again, in ADBC, we have BD = AD, DC = AD
DB = DC |Given] ...()
2DBC= DCB ...i) [Sides opposite to equal angles of a triangle are equal]
Thus, BC= BD + DC = AD + AD
(Angles opposite toequal sides of a triangle are equal] BC = 2AD
[From (i))
Subtracting (ii) from (i), we get
6
11. We have given, AABC is an isosceles triangle in and AB= BC [Sides of a
which AC=BC. AD and BE are two altitudes to sicdes BC From (i) and (ii), we get square]...i)
and AC, respectively. OB = BC
Now, in AABC, AC = BC [Given]
LABC = ZCAB So, AOCB is an isosceles triangle.
[Angles opposite to equal sides As, ZOBA = 60 [Angle of arn equilateral triangle]
of a triangle are equall
LABD = ZEAB So,ZOBC- 2CBA - 20BA =90° - 60° : zCBA = 9ne
..) ZOBC = 30°
In AAEB and ABDA,
ZAEB= ZBDA Each equals 90° as ZBOC =/0CB =75 (: 0B= BCI
D
AD I BCand BE lAC]
ZEAB = ZDBA Thus, ZOCD = 90°-75° = 15°
[From (1)]
AB = BA [Common] Similarly, LODC = 15º
AAEB ABDA (By AAS congruence] From (iii) and (iv),we have ZODC = Z0CD = 15°
...iu)
AE = BD [By CP.C.TJ OD = 0C
12. In AXYZ, XY = XZ AOCD is an isosceles triangle.
LXYA = ZXZB ...) 14. We have given, the bisectors of ZB and ZC of an
(Angles opposite to equal sides of atriangle are equal]
In AXYA and AXZB isosceles triangle ABC with AB = AC, intersect each other
ZYXA= LZXB at O.
[Given
XY = XZ [Given] Now in AABC, AB = AC [Given]
XYA= XZB [From (1)] ZACB = ZABC
AXYA AXZB [By ASA congruence] [Angles opposite to equal
XA = XB [By C.P.C.TJ sides of a triangle are equal] 3
Now, in AXAB 1
XA = XB (Proved above] LACB=ZABC
AXAB is an isosceles triangle.
13. Given: ABCD is a square DD 5 Z0CB = 20BC ..)
75
152C
75° Since, OB and OC are the bisectors of ZB and 2C
and O is a point in the interior
of it such that OAB is an respectively]
equilateral triangle. ZMOC = /0BC + 20CB
To prove : AOCD is an [Exterior angle of a triangle is equal
isosceles triangle. to the sum of two opposite interior angles]
Proof : Given, AB = 0A = OB 30 LMOC = 20BC + Z0BC [From ()]
[: AOAB is equilateral] A60° 60°B ZMOC = 2/0BC
..) ZMOC = LABC [ OB is the bisector of ZB)
 7

Chapter Test

SECTION -A 8 Two equilateral triangles are congruent

when their are equal.
Mutiple Choice Type Questions
VSA Type Questions
1. Which of the following is not a congruence 9. Write ASA congruence rule for two triangles.
criterion for triangles?
(a) SSS (b) RHS 10. Is it possible to construct a triangle with
(c) AAA (d) SAS lengths of its sides as9 cm, 7 cm and 17 cm? Give
2. In APQR, ZR =ZP and QR = 4 cm and reason for your answer.
PR= 5cm. Then, the length of PQ is Case Study-Based Questions
(a) 4 cm (b) 5 cm
(d) 2.5 cm
Attempt any 4 sub parts from each question.
(c) 2 cm Each sub-part carries I mark.
3. In a AABC, AB = 5cm, AC = 5 cm and ZA 11. Ankit is very much worried about his maths
=50°, then B=
(d) 40°
exam as he don't know how to prove that sum of
(a) 35° (b) 65° (c) 80
all three angles of a triangle is 180°. He went to
4. In figure, the measure ofZB'A'C'is his Aunt for help, who is a maths expert, she told
A Ankit to draw a triangle ABC on his notebook
and extend the side BC to D. Then she told him
3x 2x+20 to draw CE || AB. Further, he was told to name,
6 cm 6 cm
1to 5 to different angles as shown in figure.
60
5 cm B 5 cm

(a) 50° (b) 60° (c) 70° (d) 80°

Fill in the Blanks

D. If AABC ADEE. then EDF = and
LBCA =
Angles oppositeto egual sides of an isosceles
Then, she ask the following questions to Ankit.
triangle are then A
Answer them.
n AABC, BC = AB and ZB = 100°, () As BA || CE and AC is a transversal line,
1s equal to then show that Z1 is equal to 24.
(ii) What is the value of 23 + 24+ 25?
(iii)Find the value of 24 + 25.
(iv) What is the value of 1+ 22+ 23?
A100°
8

35

B
(v) IfAB= AC, and ZACD= 117°, then find the
value of ZB. 15. Line-segment AB is parallel to another line.
12. On her birthday, Priyanshi's mother gave segment CD. O
is the mid-point of AD. Show that
her beautiful diamond ring as shown in the (i) AAOB ADOC (ii) 0 is also the mid-point of
figure. Priyanshi draw the shape of ring on plane BC.
paper and marked the angle and vertices as given C
below, where T is the mid-point of PQ.
S R
Jo
B

16. In the given figure, ABCLD is a square and

pA Mis the mid-point of AB. PQ L CM meets AD at
On seeing the diagram her brother ask many Pand CBproduced at Q. Prove that PA = BQ.
D C
questions that come into his mind. Answer them.
i) ZRTP is equal to
(a) RPT (b) ZSTQ
(c) ZSQT (a) <QsT
B
(ii) Length of PT is equal to M
(a) TQ (b) TR
(c) TS (d) PR
(iii) APTR and AQTS are congruent by which 17. In the given figure, find the length of PM.
Congruency criterion?
(a) SAS (b) SSS
(c) ASA (d) RHS
(iv) Length of PR is equal to 5 cm

(a) SQ (b) SR
(C) SP (a) PQ M
(v) The medians of an equilateral triangle are
OR
(a) equal to each other
(b) not equal CDE is an equilateral triangle formed on a side
(c) equal to side of the triangle. CDof a square ABCD. Show that AADE ABCE.
(d) None of these.

SECTION -B
13. In figure, Q= ZR, QS = TR and Z1 = 22.
Show that APQT APRS.

R SECTIONC
18. ABCis an isosceles triangle with AB = AC
14. In figure, it is given that AE = AD and and BD and CE are its two medians, Show that
BD = CE.Prove that AAEB AADC. BD = CE.
9

10 In ADEF, ZE=2 ZF. DM isthe angle bisector

of /EDF that intersects EF at M. If DM = ME.
then find the value of EDE
20. Vandana wishes to literate the poor children
of the nearby slum area. She makes flash cards
for them as shown in the given figure. 23. Two lines l and m intersect at the point O
F and P is a point on a line n passing through the
point O such that P is equidistant from l and m.
3cm 3.5 cm 3 cmy 3 cm Prove that n is the bisector of the angle formed
70° by land m.
B D 3.5 cm E P
70°0
3.5 cm
(a) (b) (C)
SECTION -D
i) Which two flash cards arecongruent? 24. Bisectors of the angles B and C of an isosceles
(i) Which criteria of congruency is satisfied triangle ABC with AB = AC intersect each other
here? at O. Show that external angle adjacent to ZABC
(ii) Write the third side of both the triangles is equal to ZBOC.
which are equal. 25. Prove that the medians of an equilateral
21. ABC is an isosceles triangle with AB = AC triangle are equal.
and BD and CE are its two medians. Show that 26. If the bisector of an angle of a triangle also
BD = CE. bisects the opposite side, prove that the triangle
is isosceles.
22. In the given figure, P and Q are points on
side MN of a ALMN such that MP= QN and LP 27. In APQR, if PQ = QR and L, M and N are
= LQ. Show that ALMP ALNQ. respectively the mid-points of the sides PQ, QR
and RP. Prove that LN= MN.
28. If AABC is a right triangle with right
at Bsuch that ZBCA = 2 ZBAC. Show angle
that
hypotenuse AC = 2BC.
29. O is a point in the interior of a
M
square ABCD
such that OAB is an equilateral triangle.
P
that AOCD is an isosceles triangle. Show
OR
In AABC, AB = AC and 1 = 22. Prove that
ZPBC = PCB.