Download CONGRUENCE OF TRIANGLES

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CONGRUENCE OF TRIANGLES
1. Two figures are congruent, if they are of the same shape and of the same size.
2. Two circles of the same radii are congruent.
3. Two squares of the same sides are congruent.
4. If two triangles ABC and PQR are congruent under the correspondence A – P,
B-Q and C-R, then symbolically, it is expressed as Δ ABC  Δ PQR.
5. SAS Congruence Rule: If two sides and the included angle of one triangle are
equal to two sides and the included angle of the other triangle, then the two
triangles are congruent. (Axiom: This result cannot be proved with the help of
previously known results.)
6. ASA Congruence Rule: If two angles and the included side of one triangle are
equal to two angles and the included side of the other triangle, then the two
triangles are congruent (ASA Congruence Rule).
Construction: Two triangles are given as follows, where ABC  DEF
ACB  DFE . Sides AB=DE
To Prove: ABC  DEF
and
Proof: ABC  DEF (given)
AB = DE
AC= DF (Sides opposite to corresponding angles are in the same ratio as ratio
of angles)
Hence, by SAS congruence rule ABC  DEF is proved.
A
B
C
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D
E
F
7. AAS Congruence Rule: If two angles and one side of one triangle are equal to
two angles and the corresponding side of the other triangle, then the two triangles
are congruent.
This theorem can be proved in similar way as the previous one.
8. Angles opposite to equal sides of a triangle are equal.
9. Sides opposite to equal angles of a triangle are equal.
10. Each angle of an equilateral triangle is of 60°.
11. SSS Congruence Rule: If three sides of one triangle are equal to three sides of
the other triangle, then the two triangles are congruent.
12. RHS Congruence Rule: If in two right triangles, hypotenuse and one side of a
triangle are equal to the hypotenuse and one side of other triangle, then the two
triangles are congruent (RHS Congruence Rule).
13. In a triangle, angle opposite to the longer side is larger (greater).
14. In a triangle, side opposite to the larger (greater) angle is longer.
15. Sum of any two sides of a triangle is greater than the third side.
Theorem: Angles opposite to equal sides of an isosceles triangle are equal.
Theorem: The sides opposite to equal angles of a triangle are equal.
Theorem: If two sides of a triangle are unequal, the angle opposite to the longer
side is larger (or greater).
Theorem: In any triangle, the side opposite to the larger (greater) angle is longer.
Theorem: The sum of any two sides of a triangle is greater than the third side.
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EXERCISE 1
C
1. In quadrilateral ACBD, AC = AD and AB bisects
 A. Show that Δ ABC  Δ ABD.
Answer: In ACB & ADB
AC=AD
B
A
CAB  DAB (AB is bisecting CAD )
AB = AB (common side in both triangles)
D
So, by SAS axiom it is proved that;
ACB  ADB
2. ABCD is a quadrilateral in which AD = BC and
 DAB =  CBA . Prove that
(i) Δ ABD  Δ BAC
(ii) BD = AC
(iii)  ABD =  BAC.
D
A
Answer:
(i) In ABD & BAC
AD=BC
AB=AB (common)
B
C
BAD  ABC
So, by SAS rule ABD  BAC
(ii) Since, ABD  BAC ,
so BD = AC
(third corresponding sides of respective triangles).
B
C
(iii) In congruent triangles all corresponding angles
Are always equal, so BAD  ABC is proved
3. AD and BC are equal perpendiculars to a line
segment AB. Show that CD bisects AB.
O
D
Answer: In BOC & AOD
BC=AD (given)
CBO  DAO (Right Angle)
BOC  AOD (Opposite angles of intersecting lines
So, by ASA rule BOC  AOD
 BO  AO and it is proved that CD bisects AB.
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4. l and m are two parallel lines intersected by
another pair of parallel lines p and q. Show that
Δ ABC  Δ CDA.
p
q
A
Answer: In ABC & CDA
AB = CD (l and m are parallel)
AD=BC (AB and CD are parallel)
D
B
ABC  DCm
l
m
C
(Angles on the same side of transversal BC)
DCm  ADC (Alternate Angles are equal)
So, ABC  ADC
So, by SAS rule ABC  CDA
5. Line l is the bisector of an angle A and B is any
point on l. BP and BQ are perpendiculars from B
to the arms of A . Show that:
(i) Δ APB  Δ AQB
(ii) BP = BQ or B is equidistant from the arms
of A .
Q
l
B
A
P
Answer: In APB & AQB
AB=AB (Common side)
PAB  QAB (AB is bisector of QAP )
AQB  APB (Right Angle)
So, by ASA rule APB  AQB
E
And BQ=BP
A
6. In the given figure, AC = AE, AB = AD and
 BAD =  EAC. Show that BC = DE.
Answer: In ABC & ADE
AB=AD (given)
AC=AE (given)
Since, BAD  EAC
B
D
C
So, BAD  DAC  EAC  DAC
Or, BAC  DAE
So, by SAS rule ABC  ADE
D
E
 BC  DE proved
7. AB is a line segment and P is its mid-point. D and
E are points on the same side of AB such that
 BAD =  ABE and  EPA =  DPB
A
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P
B
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Show that
(i) Δ DAP  Δ EBP
(ii) AD = BE
DAP  EBP
BAD  ABE (given)
EPA  DPB (given)
So, EPA  EPD  DPB  EPD
Or, DPA  EPB
Answer: In
AP=PB (Since P is mid point)
So, by ASA rule DAP  EBP
So, AD=BE
A
D
M
B
8. In right triangle ABC, right angled at C, M is
the mid-point of hypotenuse AB. C is joined
to M and produced to a point D such that
DM = CM. Point D is joined to point B.
Show that:
(i) Δ AMC  Δ BMD
(ii)  DBC is a right angle.
(iii) Δ DBC  Δ ACB
(iv) CM =
1
AB
2
Answer: In AMC & BMD
BM=AM (M is midpoint)
DM=CM (given)
DMB  AMC (opposite angles)
So, AMC  BMD
Hence, DB=AC
DBA  BAC
So, DB||AC (alternate angles are equal)
So, BDC  ACB = Right Angle)
(internal angles are complementary in
Case of transversal of parallel lines)
DBC & ACB
DB=AC (proved earlier)
BC=BC (Common side)
BDC  ACB (proved earlier)
So, DBC  ACB
So, AB=DC
So, AM=BM=CM=DM
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So, CM=
1
AB
2
EXERCISE 2
1. In an isosceles triangle ABC, with AB = AC,
the bisectors of  B and  C intersect
each other at O. Join A to O. Show that :
(i) OB = OC (ii) AO bisects  A
A
Answer: In OBC
OBC  OAC (they are half of angles B & C)
O
So, OB=OC ( Sides opposite to equal angles)
In AOB & AOC
AB=AC (given)
OB=OC (proved earlier)
C
B
ABO  ACO (they are half of angles B & C)
So, AOB  AOC (SAS Rule)
So, BAO  CAO
It means that AO bisects A
A
2. In Δ ABC, AD is the perpendicular bisector of BC.
Show that Δ ABC is an isosceles triangle in which
AB = AC.
Answer: In ABD & ACD
AD=AD (common side)
BD=CD (given)
B
C
D
ADB  ADC (right angle)
So, ABD  ACD
A
So, AB=AC, which proves that ABC is isosceles
3. ABC is an isosceles triangle in which altitudes
BE and CF are drawn to equal sides AC and AB
respectively. Show that these altitudes are equal.
Answer: In ABE & ACF
AB=AC (given)
F
B
BAE  CAF (common to both triangles)
CFA  BEA (right angles)
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So, ABE  ACF ( ASA Rule )
So, BE=CF
4. ABC is a triangle in which altitudes BE and CF to
sides AC and AB are equal. Show that
(i) Δ ABE  Δ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Answer: This can be solved like previous question.
5. ABC and DBC are two isosceles triangles on the same base BC. Show that
=  ACD.
 ABD
A
B
C
D
D
Answer: ABC  ACB
DBC  DCB
So, ABC  DBC  ACB  DCB
 ABD  ACD
6. ΔABC is an isosceles triangle in which AB = AC.
Side BA is produced to D such that AD = AB.
Show that  BCD is a right angle.
A
C
Answer: In ADC & ABC
AD=AB
AC=AC
ACB  ABC
ACD  ADC
In ΔABC, ACB  ABC  CAB  180
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 CAB  180  2ACB ------------------------------ (1)
Similarly in ΔADC, DAC  180  2ACD ------------(2)
As BD is a straight line, so CAB  DAC  180
So, adding equations (1)&(2) we get
180  360  2ACB  2ACD
 180  360  2(ACB  ACD )
 2(ACB  ACD )  180
 ACB  ACD  BCD  90
7. ABC is a right angled triangle in which
and AB = AC. Find  B and  C.
 A = 90°
Answer: If AB= AC then angles opposite to these sides will be equal. As you know
the sum of all angles of a triangle is equal to 180°,
So,  A+  B+  C=180°
Or, 90°+  B+  C=180°
Or,  B+  C=180°-90°=90°
Or,  B=  C=90°
8. Show that the angles of an equilateral triangle are 60° each.
Answer: As angles opposite to equal sides of a triangle are always equal. So, in case
of equilateral triangle all angles will be equal. So they will measure one third of 180°,
which is equal to 60°
EXERCISE 3
1. Δ ABC and Δ DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC . If AD is extended
to intersect BC at P, show that
(i) Δ ABD  Δ ACD
(ii) Δ ABP  Δ ACP
(iii) AP bisects  A as well as  D.
(iv) AP is the perpendicular bisector of BC.
Answer: In ABD & ACD
AB=AC
BD=CD
AD=AD
So, Δ ABD  Δ ACD (SSS Rule)
A
D
C
B
In ABP & ACP
AB=AC
AP=AP
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ABP  ACP (Angle opposite to equal sides)
So, Δ ABP  Δ ACP (SAS Rule)
Since Δ ABP  Δ ACP
So, BAP  CAP
So, AP is bisecting BAC
Similarly BDP & CDP can be proved to be
A
P
Congruent and as a result it can be proved that
AP is bisecting BDC
2. AD is an altitude of an isosceles triangle ABC in
which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects  A.
B
M
C
Q
R
N
Answer: This can be solved like previous question.
3. Two sides AB and BC and median AM
of one triangle ABC are respectively
equal to sides PQ and QR and median
PN of Δ PQR. Show that:
(i) Δ ABM  Δ PQN
(ii) Δ ABC  Δ PQR
A
Answer: In ABM & PQN
AB=PQ
AM=PN
BM=QN (median bisects the base)
ABM  PQN
In ABC & PQR
F
E
So,
AB=PQ
BC=QR
AC=PR (Equal medians means third side will be equal)
C
B
So, ABC  PQR
4. BE and CF are two equal altitudes of a triangle ABC.
Using RHS congruence rule, prove that the triangle
ABC is isosceles.
A
Answer: In AEB & AFC
BE=CF (Perpendicular)
AB=BC (Hypotenuse)
So, AEB  AFC
5. ABC is an isosceles triangle with AB = AC.
Draw AD  BC to show that  B =  C.
Answer: After drawing AD
In ADC & ADB
 BC
B
AC=AB
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C
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AD=AD
ADC  ADB
So, ADC  ADB
So, ACD  ABC
EXERCISE 4
1. Show that in a right angled triangle, the hypotenuse
is the longest side.
A
Answer: In a right angled triangle, the angle opposite
To the hypotenuse is 90°, while other two angles are
Always less than 90°. As you know that the side
opposite to the largest angle is always the largest in
a triangle.
2. In the given triangle sides AB and AC of Δ ABC are
extended to points P and Q respectively. Also,
 PBC <  QCB. Show that AC > AB.
B
C
Q
P
Answer: ABC  180  PBC
ACB  180  OCB
Since PBC  OCB
So, ABC  ACB
As you know side opposite to the larger angle is larger
than the side opposite to the smaller angle.
Hence, AC>AB
3. In the given figure
Show that AD < BC.
D
B
 B <  A and  C <  D.
O
A
C
Answer: AO<BO (Side opposite to smaller angle)
DO<CO (Side opposite to smaller angle)
So, AO+DO<BO+CO
Or, AD<BC
D
4. AB and CD are respectively the smallest and
longest sides of a quadrilateral ABCD. Show that
 A >  C and  B >  D.
Answer: Let us draw two diagonals BD and AC as
shown in the figure.
In ΔABD Sides AB<AD<BD
So, ADB  ABD ----------------- (1)
Angle opposite to smaller side is smaller
In ΔBCD Sides BC<DC<BD
So, BDC  CBD ---------------- (2)
A
B
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Adding equation (1) & (2)
ADB  BDC  ABD  CBD
 ADC  ABC
Similarly in ABC
BAC  ACB ------------------ (3)
In ADC
DAC  DCA ----------------- (4)
Adding equations (3) & (4)
BAC  DAC  ACB  DCA
 BAD  BCD
5. In following figure, PR > PQ and PS bisects
 QPR. Prove that  PSR >  PSQ.
P
4 5
3
1
2
Q
S
R
Answer: For convenience let us name these angles as follows:
PQR  1
PRQ  2
QPR  3
QPS  4
RPS  5
PSQ  6
PSR  7
Since, PR>PQ , so
In
1  2
PQS
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1  4  6  180
In PRS
2  5  7  180
In both these triangles
4  5
1  2
So, for making the sum total equal to 180° the following will always be true:
6  7
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