congruent triangles

Made By – Nitin ChhaperwalClass 9 R.No-15

Can be classified by the number of

congruent sides

Has no congruent sides

Has at least two congruent sides

Has three congruent sides

Can be classified by the angle

measures

Has one right angle

Has three acute angles

Triangle with one obtuse angle

Cut any shape triangle out of a sheet of paper

Tear off the corners. Piece them together

by having the corners touch.

The sum of the angles of a triangle

is 180 degrees

Congruence of Triangles

Congruent figures can be rotations of one another.

Congruent figures can be reflections of one another.

A B

C

X Y

Z

∆ABC is congruent to ∆XYZ

A B

C

X Y

Z

Corresponding parts of these triangles are congruent.


A B

C

X Y

Z


Corresponding parts are angles and sides that “match.”


A B

C

X Y

Z


A X


A B

C

X Y

Z


B Y


A B

C

X Y

Z


C Z


A B

C

X Y

Z


AB XY


A B

C

X Y

Z


BC YZ


A B

C

X Y

Z


AC XZ

∆DEF is congruent to ∆QRS

D E

F

Q

R

S


D E

F

Q

R

S



D E

F

Q

R

S


D Q


D E

F

Q

R

S


E R


D E

F

Q

R

S


F S


D E

F

Q

R

S


DE QR


D E

F

Q

R

S


DF QS


D E

F

Q

R

S


FE SR

1. SIDE – ANGLE – SIDE RULE (SAS RULE)

Two triangles are congruent if any two sides and the includes angle of one triangle is equal to the two sides and the included angle of other triangle.

EXAMPLE :- (in fig 1.3)

GIVEN: AB=DE, BC=EF ,

B= E

SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule

▲ABS = ▲DEF

4 cm4 cm

600 600

A

B C

D

FE

Fig. 1.3

2. ANGLE – SIDE – ANGLE RULE (ASA RULE )

Two triangles are congruent if any two angles and the included side of one triangle is equal to the two angles and the included side of the other triangle.

EXAMPLE : (in fig. 1.4)

GIVEN: ABC= DEF,

ACB= DFE,

BC = EF

TO PROVE : ▲ABC = ▲DEF

ABC = DEF, (GIVEN)

ACB = DFE, (GIVEN)

BS = EF (GIVEN)

▲ABC = ▲DEF (BY ASA RULE)

A

B C

D

E F

Fig. 1.4

3. ANGLE – ANGLE – SIDE RULE (AAS RULE)

Two triangles are congruent if two angles and a side of one triangle is equal to the two angles and one a side of the other.

EXAMPLE: (in fig. 1.5)

GIVEN: IN ▲ ABC & ▲DEF

B = E

A= D

BC = EF

TO PROVE :▲ABC = ▲DEF

B = E

A = D

BC = EF

▲ABC = ▲DEF (BY AAS RULE)

D

E F

A

B C

Fig. 1.5

4. SIDE – SIDE – SIDE RULE (SSS RULE)

Two triangles are congruent if all the three sides of one triangle are equal to the three sides of other triangle.

Example: (in fig. 1.6)

Given: IN ▲ ABC & ▲DEF

AB = DE , BC = EF , AC = DF


AB = DE (GIVEN )

BC = EF (GIVEN )

AC = DF (GIVEN )

▲ABC = ▲DEF (BY SSS RULE)

D

E F

A

B C

Fig. 1.6

5. RIGHT – HYPOTENUSE – SIDE RULE (RHS RULE )

Two triangles are congruent if the hypotenuse and the side of one triangle are equal to the hypotenuse and the side of other triangle.

EXAMPLE : (in fig 1.7)

GIVEN: IN ▲ ABC & ▲DEF

B = E = 900 , AC = DF , AB = DE


B = E = 900 (GIVEN)

AC = DF (GIVEN)

AB = DE (GIVEN)

▲ABC = ▲DEF (BY RHS RULE)

D

E F

A

B C

900

900

Fig. 1.7

1. The angles opposite to equal sides are always equal.

Example: (in fig 1.8)

Given: ▲ABC is an isosceles triangle in which AB = AC

TO PROVE: B = C

CONSTRUCTION : Draw AD bisector of BAC which meets BC at D

PROOF: IN ▲ABC & ▲ACD

AB = AD (GIVEN)

BAD = CAD (GIVEN)

AD = AD (COMMON)

▲ABD = ▲ ACD (BY SAS RULE)

B = C (BY CPCT)

A

B D C

Fig. 1.8

2. The sides opposite to equal angles of a triangle are always equal.

Example : (in fig. 1.9)

Given : ▲ ABC is an isosceles triangle in which B = C

TO PROVE: AB = AC

CONSTRUCTION : Draw AD the bisector of BAC which meets BC at D

Proof : IN ▲ ABD & ▲ ACD

B = C (GIVEN)

AD = AD (GIVEN)

BAD = CAD (GIVEN)

▲ ABD = ▲ ACD (BY ASA RULE)

AB = AC (BY CPCT)

A

B D C

Fig. 1.9

When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation.

Theorem 1 . If two sides of a triangle are unequal the larger side has the greater angle opposite to it. Example: (in fig. 2.1)

Given : IN ▲ABC , AB>AC

TO PROVE : C = B

Draw a line segment CD from vertex such that AC = AD

Proof : IN ▲ACD , AC = AD

ACD = ADC --- (1)

But ADC is an exterior angle of ▲BDC

ADC > B --- (2)

From (1) &(2)

ACD > B --- (3)

ACB > ACD ---4

From (3) & (4)

ACB > ACD > B , ACB > B ,

C > B

A

B

D

C

Fig. 2.1

THEOREM 2. In a triangle the greater angle has a large side opposite to it

Example: (in fig. 2.2)

Given: IN ▲ ABC B > C

TO PROVE : AC > AB

PROOF : We have the three possibility for sides AB and AC of ▲ABC

(i) AC = AB

If AC = AB then opposite angles of the equal sides are equal than

B = C

AC ≠ AB

(ii) If AC < AB

We know that larger side has greater angles opposite to it.

AC < AB , C > B

AC is not greater then AB

(iii) If AC > AB

We have left only this possibility AC > AB

A

CB

Fig. 2.2

THEOREM 3. The sum of any two angles is greater than its third side

Example (in fig. 2.3) TO PROVE : AB + BC > AC

BC + AC > AB

AC + AB > BC

CONSTRUCTION: Produce BA to D such that AD + AC .

Proof: AD = AC (GIVEN)

ACD = ADC (Angles opposite to equal sides are equal )

ACD = ADC --- (1)

BCD > ACD ----(2)

From (1) & (2) BCD > ADC = BDC

BD > AC (Greater angles have larger opposite sides )

BA + AD > BC ( BD = BA + AD)

BA + AC > BC (By construction)

AB + BC > AC

BC + AC >AB

A

CB

D

Fig. 2.3

THEOREM 4. Of all the line segments that can be drawn to a given line from an external point , the perpendicular line segment is the shortest.

Example: (in fig 2.4)

Given : A line AB and an external point. Join CD and draw CE AB

TO PROVE CE < CD

PROOF : IN ▲CED, CED = 900

THEN CDE < CED

CD < CE ( Greater angles have larger side opposite to them. )

BA

C

ED Fig. 2.4

congruent triangles

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