 ## Congurency and Similarities

Subject: Compulsory Maths

#### Overview

Two triangles are congruent if they have exactly the same three sides and exactly the same three angles. Similar triangle has the same shape, but the size may be different. Two triangles are similar if and only if the corresponding sides are in proportion and the corresponding angles are congruent.

### Congruent Triangles

The triangles having same size and shape are called congruent triangles. Two triangles are congruent when the three sides and three angles of one triangle have the measurements as three sides and three angles of another triangle. The symbol for congruent is ≅.

In the following figure, ΔABC and ΔPQR are congruent. We denote this as ΔABC ≅ ΔPQR. #### Postulate and Theorems for Congruent Triangles

Postulate (SAS)

If two sides and the angle between them in one triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent. In the given figure,

AB ≅ PQ Sides (S)

∠B ≅ ∠Q Angle (A)

BC ≅ QR Side (S)

Therefore, ΔABC ≅ ΔPQR  Theorem (ASA)

A unique triangle is formed by two angles and the included side.

Therefore, if two angles and the included side of one triangle are congruent to two angles and the included side of the another triangle, then the triangles are congruent. In the figure,

∠B ≅ ∠E Angle (A)

BC ≅ EF Side (S)

∠C ≅ ∠F Angle (A)

Therefore, ΔABC ≅ ΔDEF

Theorem ( AAS)

A unique triangle is formed by two angles and non-included side. Therefore, if two angles and the side opposite to one of them in a triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent. In the figure,

∠A ≅ ∠X Angle (A)

∠C ≅ ∠Z Angle (A)

BC ≅ YZ Side (S)

Therefore, ΔABC ≅ ΔXYZ

Theorem (SSS)

A unique triangle is formed by specifying three sides of a triangle, where the longest side (if there is one) is less than the sum of the two shorter sides.

Therefore, if their sides of a triangle are congruent to three sides of another triangle, then the triangles are congruent. In the figure

AB ≅ PQ Sides (S)

BC ≅ QR Sides (S)

CA ≅ RP Sides (S)

Therefore, ΔABC ≅ ΔPQR

### Similar Triangles

#### Methods of providing triangles similar

1. If the corresponding sides of a triangle are proportion to another triangle then the triangles are similar.
Example If∠A ≅ ∠D and ∠B ≅ ∠E, Then ΔABC ∼ ΔDEF

2. If the corresponding angle of a triangle is congruent to another triangle then, the triangles are similar.
Example If $\frac{AB}{DE}$ = $\frac{BC}{EF}$ = $\frac{AC}{DF}$, then ΔABC ∼ ΔDEF

3. Conversing first and the second method we can prove triangle similar as their sides being proportional and angles congruent.
Example ∠A ≅ ∠D and $\frac{AB}{DE}$ = $\frac{AC}{DF}$ then ΔABC ∼ ΔDEF

#### In case of overlapping triangles

When the lines are parallel in a triangle, then they intersect each other which divides the sides of a triangle proportionally.

Verification: In ΔPQR and ΔSPT

 Statements Reasons ST⁄⁄QR Given $\angle$PST $\cong$ $\angle$QSR Corresponding angles ΔPQR $\cong$ ΔSPT Common Angle P $\frac{PS}{SQ}$ = $\frac{PT}{TR}$ ST⁄⁄QR

Example

Given the following triangles, find the length of x. Solution:

The triangles are similar by AA rule.So, the ratio of lengths are equal.

$\frac{6}{3}$ = $\frac{10}{x}$

or, 6x = 30

or, x = $\frac{30}{6}$

$\therefore$ x = 5 cm

##### Things to remember
• There is three easy way to prove similarity. If two pairs of corresponding  angles in a pair of triangles are congruent, then the triangles are similar.
• When the three angle pairs are all equal, the  three pairs of the side must  be proportion.
• When triangles are congruent and one triangle is placed on the top of other sides and angles that are in the same position are called corresponding parts.
• Congruent and similar shapes can make calculations and design work easier.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

Here, AB=1.9 cm DE=1.7 cm

BC=1.7 cm EF=1.3 cm

CA=2.1 cm DF=2.1 cm

$\angle$A=107o$\angle$D=90o

$\angle$B=57o$\angle$E=35o

$\angle$C=16o$\angle$F=55o

The angles arms of the triangle is not equal so it is not congurent.

Solution:

Here, AB=1.6 cm DE=1.9 cm

BC=1.3 cm EF=1.4 cm

CA=1.9 cm DF=1.9 cm

$\angle$A=45o$\angle$D=40o

$\angle$B=100o$\angle$E=70o

$\angle$C=35o$\angle$F=70o

The angles arms of the triangle is not equal so it is not congurent.

Solution:

Here given, congurent triangle is PQ and LM, QR and MN, PR and LN. So that, congurent angles is $\angle$P and(\angle\)L,(\angle\)R and(\angle\)N,(\angle\)Q and(\angle\)M.

Solution:

Here given, congurent triangle is XY and AB, YZ and BC, XZ and AC. So that, congurent angles is $\angle$X and(\angle\)A,(\angle\)Y and(\angle\)B,(\angle\)Z and(\angle\)C.

Solution:

In $\triangle$ ABC and $\triangle$A'BC'

$\angle$BAC≈$\angle$BA'C',$\angle$ACB≈$\angle$BC'A' being corresponding angles.

so, $\triangle$ABC ~$\triangle$A'BC'

$\frac{AB}{A'B}$=$\frac{BC}{BC'}$=$\frac{AC}{A'C'}$

$\frac{30+x}{30}$=$\frac{15+y}{y}$=$\frac{22}{14'}$

Taking 1st and 3rd $\frac{30+x}{30}$=$\frac{22}{14}$

14(30+x)=22$\times$30

420+14x=660

14x=660-420

14x=240

x=$\frac{240}{14}$=17.1 cm

Taking 2nd and 3rd

$\frac{15+y}{y}$=$\frac{22}{14}$=$\frac{11}{7}$

11 y=105+7 y

11y-7y=105

4y=105

y=$\frac{105}{4}$=26.25 cm

Solution,

Here, 5y-3o=62o[ $\therefore$

or, 5y-3o=62o

or, 5y=62o+3o=65o

or, y=$\frac{65^o}{5}$=13o

Likewise, 4x-4o=48o

or, 4x-4o=48o

or, 4x=48o+4o=52o

or, x=$\frac{52^o}{4}$=13o

$\therefore$x=13o and y=13o Ans.

Solution:

We know that,

$\angle$A+$\angle$B+$\angle$C=180o

or, 122o+38o+$\angle$C=180o

or,$\angle$C=180o-160o=20o

Again,$\angle$R=$\angle$C [$\therefore$ ]

or,$\angle$R=$\angle$A

or, x=20o

Likewise,$\angle$P=$\angle$A

or, y+10o=122o

or, y=112o

and$\angle$Q=$\angle$B

or,$\angle$Q=38o

$\therefore$ x=20o and y = 112o Ans.

Solution:

Here, $\angle$A+ $\angle$B + $\angle$C[$\therefore$ sum of interior angles of a triangle is 180o]

or, 56o+$\angle$B+64o=180o

or,$\angle$B=180o-120o=60o

Again, AC=PR

or, 3x-0.6=x+3

or, 3x-x=3+0.6

or, 2x=3.6

or, x=$\frac{3.6}{2}$=1.8

$\therefore$PR=AC=(3x-0.6)cm

=(3$\times$1.8-0.6)cm

=(5.4 - 0.6) cm

=95.4 - 0.6) cm

=4.8 cm

AB=PQ=3.1 cm [ $\therefore$ corresponding sides of congurent triangles]

so,$\angle$B=60o, AC=PR=4.8 cm, AB=PQ=3.1 cm

Solution:

Here, $\angle$L+ $\angle$M+ $\angle$N [$\therefore$ sum of interior angles of a triangle is 180o]

or, 45o + 70o + $\angle$N=180o

or, $\angle$N=180o-115o=65o

Again, $\angle$X=45o and $\angle$Z=65o [$\therefore$ Corresponding angles of congurent triangle]

Likewise, MN=YZ

or, 2 y=2.5

or, y=1.25

2x+1=3

or, 2x=2

or, x=1

so that , $\angle$L= $\angle$X=45o, $\angle$N= $\angle$Z=65o, MN=YZ=2y=2.5 cm and LM=XY=3 cm, x=1, y= 1.25 Ans.

Solution:

Here, $\angle$A=52o $\angle$=88o $\angle$C=40o

$\angle$P=52o $\angle\Q=88o \(\angle$R=40o

AB=1 cm BC=1.2 cm CA=2 cm

PQ=1.3 cm QR=1.6 cm PR=2.6 cm

$\angle$A= $\angle$P, $\angle$B= $\angle$Q and $\angle$C= $\angle$R

$\frac{AB}{PQ}$=$\frac{1.0}{1.3}$,$\frac{BC}{QR}$,$\frac{1.0}{1.3}$,=$\frac{CA}{RP}$=$\frac{1.0}{1.3}$

Hence, given triangle is similar triangle.

Solution:

Here, $\angle$ A=40o$\angle$ B=70o$\angle$ C=70o

$\angle$ P=41o$\angle$ Q=62o$\angle$ R=77o

AB=1.5 cm BC=1 cm CA=1.5 cm

PQ=1.1 cm QR=1 cm RP=1.6 cm

Here,$\angle$A $\neq$$\angle$P,$\angle$B$\neq$$\angle$Q and $\angle$C$\neq$ $\angle$R

Hence, given triangle is not similar triangle.

Solution:

Here, $\angle$A=38o $\angle$B=90o $\angle$C=52o

$\angle$P=38o $\angle$Q=90o $\angle$r=52o

AB=1.6 cm BC=1.2 cm

CA=2 cm PQ=0.8 cm

QR=0.6 cm PR=1 cm

$\angle$A= $\angle$P, $\angle$B= $\angle$Q and $\angle$C= $\angle$R

and $\frac{AB}{PQ}$=$\frac{1.6}{0.8}$=2, $\frac{BC}{QR}$=$\frac{1.2}{0.6}$=2, $\frac{CA}{RP}$=$\frac{2}{1}$=2

Hence, given triangle is similar triangle.

Solution:

Here,In $\triangle$ADE and $\triangle$ABC

$\angle$DAE=$\angle$ABC

$\therefore$$\triangle$ADE~$\triangle$ABC

so, $\frac{DE}{BC}$=$\frac{AE}{AC}$

or,$\frac{DE}{7.2}$=$\frac{6+4}{6}$

or, $\frac{DE}{7.2}$=$\frac{10}{6}$

or, DE$\times$6=10$\times$7.2

or, DE=$\frac{10\times7.2}{6}$=12 cm

Again, $\angle$ACB=$\angle$CDE=30o

Hence, DE=12 cm and $\angle$ACD=30oAns.

From the given figure,

∠STU ≅ ∠SVW and TU ≅ VW

Here, ∠TSU and ∠VSW are vertical angles. Since vertical angles are congruent,

∠TSU ≅ ∠VSW.

Finally, put the three congruency statements in order. ∠STU is between ∠TSU and TU, and ∠SVW is between ∠VSW and VW in the diagram.

∠TSU ≅ ∠VSW (Angle)

∠STU ≅ ∠SVW (Angle)

TU ≅ VW (Side)

Hence, the given triangles are congurent as it forms AAS theorem.

From the given figure,

BC ≅ BH and ∠BCF≅∠BHG.

Here, ∠CBF and ∠GBH are vertical angles. Since vertical angles are congruent,

∠CBF ≅ ∠GBH.

Finally, put the three congruency statements in order. BC is between ∠BCF and ∠CBF, and BH is between ∠BHG and ∠GBH in the diagram.

∠BCF ≅ ∠BHG Angle

BC ≅ BH Side

∠CBF ≅ ∠GBH Angle

Hence, the congruent sides and angles form ASA. The triangles are congruent by the ASA Theorem.

From the figure,

∠XWY≅∠YWZ and ∠WXY≅∠WZY.

Here, the triangles share WY. By the reflexive property of congruence, WY ≅ WY.

Finally, put the three congruency statements in order. ∠WXY is between ∠XWY and WY, and ∠WZY is between ∠YWZ and WY in the diagram.

∠XWY ≅ ∠YWZ Angle

∠WXY ≅ ∠WZY Angle

WY ≅ WY Side

Hence, the congruent sides and angles form AAS. The triangles are congruent by the AAS Theorem.