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Similarity of Triangles by AAA Criteria

Triangles I - Concepts

Similar Figures

Similarity of Triangles

Basic Proportionality Theorem and its Converse

Similarity of Triangles by AAA Criteria

Similarity of Triangles by AA Criteria

Similarity of Triangles by SSS Criteria

Similarity of Triangles by SAS Criteria

Class - 10th NTSE Subjects

Concept Explanation

Similarity of Triangles by AAA Criteria

Similarity of Triangles by AAA Criteria:

Theorem : If in two triangles, the corresponding angles are equal, then the corresponding sides are proportional ( i.e., in the same ratio) and hence the triangles are similar.

Given: in $large Delta$ ABC and $large Delta$ DEF

$large dpi{120} large angle A=angle D,angle B=angle E;;and;;angle C=angle F$

To Prove :

$large frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF} ;and; hence;Delta ABCsim Delta DEF$

Construction: Cut DP = Ab and DQ = Ac and Join PQ.

Proof: In $large Delta$ ABC and $large Delta$ DPQ

AB = DP [ By Construction ]

$large angle$ A = $large angle$ D [ Given]

AC = DQ [ By Construction ]

Hence $large Delta$ ABC $large cong$ $large Delta$ DPQ [ By SAS Criteria]

$large angle$ B = $large angle$ P and $large angle$ C = $large angle$ Q [C.P. C. T]

But $large angle$ B = $large angle$ E [ Given]

Hence $large angle$ P = $large angle$ E

But they are corresponding angles when PQ and EF are straight lines and Transversal DE cuts them.

As they are equal

So the lines are parallel

Therefore by Basic Proportionality Theorem

$large frac{DP}{PE}= frac{DQ}{QF}$

Taking Reciprocal both Sides and adding 1 to both sides we get

$large frac{PE}{DP}+1= frac{QF}{DQ}+1 ; Rightarrow ;frac{PE +DP}{DP}=frac{QF+DQ}{DQ}$ $large Rightarrow frac{DE}{DP}= frac{DF}{DQ}$

Taking Reciprocal we get replacing DP with AB and DQ with AC as they are equal by construction We get

$large frac{AB}{DE}= frac{AC}{DF}$

Similary We can prove that

$large frac{AB}{DE}= frac{BC}{EF}$

And Hence

$large frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF} ;and; Therefore ;Delta ABCsim Delta DEF$

Illustration : If AB || CD , prove that $large Delta COD sim Delta BOA$

Solution: As AB || CD then

$large angle C = angle B$ [ Alternate Interior Angle ]

$large angle D = angle A$ [ Alternate Interior Angle ]

$large angle COD = angle AOB$ [ Vertically Opposite Angle ]

Therefore, $large Delta COD sim Delta BOA$ [ AAA Similarity Criterion]

Sample Questions

(More Questions for each concept available in Login)

Question : 1

In above figure, $bigtriangleup ABCsim bigtriangleup PRQ$ by __________________ criteria.
A. SAS	B. SSS	C. AAA	D. AAA
Right Option : D
View Explanation

Question : 2

In the following figure, $bigtriangleup ABMsim bigtriangleup AMC$ by which criteria?
A. SAS	B. AA	C. AAA	D. SSS
Right Option : C
View Explanation

Question : 3

In the following diagram, $bigtriangleup MBA sim bigtriangleup ABC$ by which similarity criteria?
A. SSS	B. SAS	C. AAA	D. AA
Right Option : C
View Explanation