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Similarity of Triangles by SSS 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 IMO Subjects

Concept Explanation

Similarity of Triangles by SSS Criteria

Similarity of Triangles by SSS criteria:

Theorem: If the corresponding sides of two triangles are proportional, then the triangles are similar.

Given: $large In ;Delta ABC ;and;Delta PQR$

$large frac{AB}{PQ}=frac{BC}{QR}=frac{CA}{RP}$

$large To Prove:;Delta ABCsim Delta PQR$

Construction: Cut PX = AB and PY = AC and join XY.

Proof: It is given that

$large frac{PQ}{AB}= frac{PR}{AC}$

Replacing AB with PX and AC with PY

$large frac{PQ}{PX}= frac{PR}{PY}$

Subtract 1 from both sides

$large frac{PQ}{PX}-1= frac{PR}{PY}-1;; Rightarrow ; frac{PQ-PX}{PX}=frac{PR-PY}{PY}$

$large frac{XQ}{PX}= frac{YR}{PY}Rightarrow ; XY parallel QR$ [Converse of Basic Proportionality Theorem]

Now XY || QR and PQ is the transversal

Therefore $large angle X = angle Q;;and;angle Y= angle R$ [Corresponding Angles are equal]

Now in $large Delta PQR ;and;Delta PXY$

$large angle P = angle P$ [Common ]

$large angle Q = angle X$ [Proved Above]

$large angle R = angle Y$ [Proved Above]

Therefore $large Delta PQR ;sim ;Delta PXY$

$large frac{PQ}{PX}=frac{QR}{XY}=frac{PR}{PY}$ ..................(1)

$large frac{AB}{PQ}=frac{BC}{QR}=frac{CA}{RP}$ .....................(2)

We know that AB = PX and AC= PY replacing AB and AC in Equation (2)

$large frac{PX}{PQ}=frac{BC}{QR}=frac{PY}{PR}$ .....................(3)

From (1) and (3) BC= XY

$large In ;Delta ABC ;and;Delta PXY$

AB = PX [By Construction ]

AC = PY [By Construction ]

BC = XY [ Proved Above ]

Therefore $large ;Delta ABC ;cong;Delta PXY$ [By SSS Criterion]

$large angle A =angle P$

$large angle B =angle X=angle Q$

$large angle C =angle Y=angle R$

Hence $large Delta ABCsim Delta PQR$

Illustration : Find the value of $large angle$ E.

Solution: $large In; Delta ABC; and; Delta DEF$

$large frac{AB}{DF}=frac{3.6}{7.2}=frac{1}{2}$

$large frac{BC}{EF}=frac{6.1}{12.2}=frac{1}{2}$

$large frac{AC}{DE}=frac{4.2}{8.4}=frac{1}{2}$

That is $large frac{AB}{DF}=frac{BC}{EF}=frac{AC}{DE}=frac{1}{2}$

So $large Delta ABC sim;Delta DFE$

Therefore $large angle C =angle E$

$large In;Delta ABC;;angle A+angle B+angle C= 180^0$

$large 60^0+80^0+angle C= 180^0Rightarrow angle C= 180^0-140^0=40^0$

$large angle E=40^0$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Check whether the two triangles are similar or not. If so, then write the similarity criteria.
A. AAA	B. AA	C. SAS	D. SSS
Right Option : D
View Explanation

Question : 2

If the above triangles are similar, then write the similarity criteria.
A. yes, SAS	B. not similar	C. yes, AAA	D. yes, SSS
Right Option : B
View Explanation

Question : 3

Whether the above triangles shown are similar or not? If so, write the similarity criteria.
A. yes, SSS	B. yes, SAS	C. not similar	D. yes, AAA
Right Option : C
View Explanation