Criteria for Similarity of Triangles

Things are often referred similar when the physical structure or patterns they show have similar properties, Sometimes two objects may vary in size but because of their physical similarities, they are called similar objects. For example, a bigger Square will always be similar to a smaller square. In Triangles, if the sizes vary but the shapes they have are alike, then the Triangles can be called Similar Triangles. When two triangles are declared Similar, their corresponding angles are always congruent (Identical in form), and their sides are Proportional. Let's learn more about the similarities of triangles.

Similar Triangles

Two or more than two triangles are similar when their corresponding angles are congruent and their sides are proportional to each other(the ratio of their sides is equal). Noticing things from our daily life or a good look in our textbooks tell us that there are so many objects similar to each other we have never noticed, For Instance, a baby bear looks like its mother even though the mother is relatively bigger, the same concept is applied here in triangles.

Difference between Similar Triangles and Congruent Triangles

• In Similar Triangles, Shape is always same, but Size may vary.
• In Congruent Triangles, Both shape and size are equal.

Thales Theorem or Basic Proportionality Theorem

A Famous Greek Mathematics “Thales” gave a universal Theorem for Triangles, “In two Triangles that are Equiangular(Corresponding angles of both triangles are equal), the Ratio of their corresponding sides are always equal.”

As we can notice, the above given Two triangles are equiangular.

∠A = ∠A [Common Angle]

∠B = ∠D [Corresponding Angles]

∠C = ∠E [Corresponding Angles]

Therefore, it is concluded "If a line is drawn parallel to one side of the triangle to intersect on remaining two sides, it will divide the remaining two sides in the same ratio”

\frac{AD}{AB}=\frac{AE}{AC}  , DF//BC

Proof of Thales Theorem

If a line is drawn parallel to one side of a triangle, intersecting other sides at distinct points, then the division of the other two sides is in the same ratio

To Prove: 

\frac{AD}{DB}=\frac{AE}{EC}

Given: 

In ΔABC, DE is parallel to BC

Construction:

Join BE and CD and draw perpendicular to AC and AB from D and E.

Proof:

In ΔADE, Area of triangle = 1/2 × DE × EN ⇢ 1

Also, 1/2 × DE× DM ⇢ 2

In ΔBDE, Area of Triangle =1/2 × BD × EN ⇢ 3

In ΔDEC, Area of Triangle= 1/2 × EC × DM ⇢ 4

Dividing equation 1 and 3

\frac{area ADE}{area BDE}=\frac{AD}{DB}    ⇢ 5

Dividing equation 2 and 4

\\frac{area ADE}{area BDE}= \frac{AE}{EC}    ⇢ 6

We know, area of ΔBDE and ΔDEC are equal, since BD is parallel to DE and they both have the same base DE

Now, from equation 5 and 6

\frac{AD}{DB}=\frac{AE}{EC}

Converse of Thales Theorem 

If a line divides any two sides of the triangle in the same ratio, then the line is parallel to the third side of the triangle

Proof:

Given: 

\frac{PM}{MQ}=\frac{PN}{NR}

Construction:

Lets assume that line MN is not parallel to QR, Draw another line MK such that MK is parallel to QR

Now, in ΔPQR, 

\frac{PM}{MQ}=\frac{PK}{KR}    ⇢ 1

but, this is also provided that,

\frac{PM}{MQ}=\frac{PN}{NR}    ⇢ 2

\frac{PK}{KR}=\frac{PN}{NR}

But, K and N lie on the same line PR, Therefore, they coincide each other and are just the same point.

Therefore, MN and MK are the same line and MN is parallel to QR.

Criteria for Similarity of Triangles

Two triangles are called similar when their angles are equal and their corresponding sides are always in the same ratio, this is what we have learned so far, however one does not need to prove all the things mentioned above to show similarity of two triangles. There are 3 easier ways to find out if the two triangles are similar.

Theorem 1: SSS Similarity Criterion

It states if in a triangle, all the sides are proportional to the sides of other triangles, then the corresponding angles will always be equal and hence, both triangles are Similar.

Proof:

To Prove: \frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}

Given: ∠A=∠D, ∠B=∠E, ∠C=∠F

Construction: Draw a line PQ in ΔDEF such that AB = DP, AC = DQ, BC = PQ

Proof: \frac{AB}{DE}=\frac{AC}{DF}

\frac{DP}{DE}=\frac{DQ}{DF}

Reciprocal the fraction, \frac{DE}{DP}=\frac{DF}{DQ}

Subtract equation by 1 on both sides\frac{DE}{DP}-1=\frac{DF}{DQ}-1

\frac{PE}{DP}=\frac{QF}{DQ}

By the converse of BPT, PQ is parallel to EF

Therefore, ∠P = ∠E, ∠Q = ∠F (by corresponding angles)

Therefore, ΔDEF∼ ΔDPQ

\frac{DP}{DE}=\frac{DQ}{DF}=\frac{PQ}{EF}

\frac{PQ}{EF}=\frac{BC}{EF}

BC=PQ, AB=DP, AC=DQ

ΔABC≅ ΔDPQ

∠A= ∠D, ∠B= ∠P, ∠C= ∠Q, ∠B= ∠E, ∠C= ∠F,

Hence, ΔABC ∼ ΔDEF

Theorem 2: AAA OR AA Similarity Criterion 

AAA referring to angles (all three) of the triangles. It states “If the two corresponding angles both the triangles are equal, then their respective sides will always have same ratio and the triangles are similar triangles”

AA is another name for the theorem since having two angles equal will automatically make the third angle of the triangles equal as the sum is always 180°.

Proof:

Given: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

To Prove: \frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}

Construction: Make a line PQ in ΔDEF such that AB=DP, AC=DQ, BC=PQ

Proof: In ΔABC and ΔDPQ 

AB=DP, AC=DQ, ∠A=∠D

By S.A.S Property of triangles, both the triangles are congruent to each other.

ΔABC≅ ΔDPQ

We can say, ∠B= ∠DPQ (By CPCT)

 ∠B= ∠E, Therefore ∠E = ∠DPQ 

By corresponding angles, it can be concluded that PQ is parallel to EF

With the help of Converse of Thales theorem,

\frac{DP}{PE}=\frac{DQ}{QF}=\frac{PQ}{EF}

\frac{PE}{DP}=\frac{QF}{DQ}

Adding 1 to both LHS and RHS

\frac{PE+DP}{DP}=\frac{QF+DQ}{DQ}

\frac{DE}{DP}=\frac{QF}{DQ}

Replace DP with AB and DQ with AC

\frac{AB}{DE}=\frac{AC}{DF}

Therefore, the ratio of corresponding sides is coming out to be equal as well.

Hence, ΔABC= ΔDEF

Theorem 3: SAS Similarity Criterion

If the two angles of both the triangles along with both the sides connected to them are equal, then the triangles are called Similar Triangles.

Proof 

To Prove: ΔABC∼ ΔDEF

Given: ∠A= ∠D, \frac{AB}{DE}=\frac{AC}{DF}

Construction: Draw a line PQ in triangle ΔDEF Such that AB= DP, AC=DQ

\frac{AB}{DE}=\frac{AC}{DF}

\frac{DP}{DE}=\frac{DQ}{DF}

Reciprocate the fraction, \frac{DE}{DP}=\frac{DF}{DQ}

Subtracting both sides by 1, \frac{PE}{DE}=\frac{QF}{DF}

So, PQ is parallel to EF (by converse of BPT)

∠P= ∠E, ∠Q= ∠F (by corresponding angles)

ΔABC= ΔDPQ

∠A= ∠D, ∠B=∠P, ∠C=∠Q

Since ∠P=∠E, ∠Q= ∠F

Therefore, ∠A=∠D, ∠B= ∠E, ∠C= ∠F

ΔABC∼ ΔDEF

Pythagoras Theorem

In a right-angled Triangle, the square of the hypotenuse is equal to the sum of squares of other two sides

Proof:

Given: ΔPQR is a right angled Triangle and QS is perpendicular to PR

Proof:

We know that, ∠PQR= ∠QSR

∠P= ∠P

Therefore, Δ PQR∼ ΔQSR

\frac{PQ}{QS}=\frac{QR}{RS}=\frac{PR}{QR}

QR²= PR× RS ⇢ 1 

Apply the same for Δ PQR and Δ PSQ

PQ²= PR× PS ⇢ 2

From 1 and 2,

PQ² + QR²= PR× RS + PR×  PS

PQ²+ QR²= PR (RS+ PS)

PQ²+ QR²= PR²

Hence, Proved.

Sample Problems

Question 1: In the figure given below, XY is parallel to BC, AX=2cm, XB= 3cm, and the base of the triangle BC= 5cm, Then find the value of XY using Thales Theorem.

Solution:

According to Thales theorem,

\frac{AX}{AB}=\frac{XY}{BC}     

\frac{2}{5}=\frac{XY}{5}

{XY}=\frac{10}{5}

XY= 2cm

Question 2: In the right-angled triangle shown below, what is the value of p in terms of q.

Solution:

ABC is a right-angled triangle,

Using Pythagoras theorem, we get,

AC²= AB²+ BC²

P²= (4q)²+ (3q)²

P²= 16q+ 9q

p²= 25q

p= 5q

Question 3: In the below-given figure, when PQ is parallel to BC, find the value of x.

Solution:

In the Triangle, PQ is parallel to BC, therefore, Thales theorem can be applied,

\frac{AP}{PB}=\frac{AQ}{QC}

\frac{2}{x+2}=\frac{5}{3x-3}

(3x-3)(2)= (x+2)(5)

6x-6= 5x+ 10

x = 16

Question 4: What are the three ways of proving the similarity of Two Triangles?

Answer:

The three ways of Proving Similarity of Triangles are:

1. AAA similarity criterion (angle-angle-angle)
2. SAS Similarity criterion (side- angle- side)
3. SSS similarity criterion (side- side- side)

Question 5: In Triangle ABC, Line DE is drawn in such a way that ∠ABC = ∠DEC, Prove that ΔABC≅ ΔDEC.

Solution: 

In ΔABC and ΔDEC, 

It is already given that ∠ABC= ∠DEC

And since, angle C is common in both the Triangles, we can say, 

∠ACB= ∠DCE

As two angles are equal, the third angle will automatically be equal since the sum of the three angles of a triangle is always 180°

Hence, from AAA Similarity Criterion, it can be concluded,

ΔABC ≅ ΔDEC

Question 6: In a Right-angled isosceles triangle, the base of the triangle is 2cm, Find the hypotenuse of the Triangle.

Solution:

The Triangle given in the question is a right-angled isosceles triangle and shall look something like this.

BC= 2cm

As it is an Isosceles triangle, AB= 2cm, 

According to Pythagoras theorem, AC²= AB² + BC²

AC²= 2²+ 2²

AC= √8 = 2√2cm

Question 7: How is SAS and SSS Criterion different from each other?

Answer:

Both Criteria have same result, that is, they both proved the triangles to be Congruent to each other, but the method of proving them is very different. In SSS criterion, when all the three sides are known to be equal, then the two Triangles are Congruent in nature. In SAS criterion, when any two sides and the angles between those two sides are equal, then the Triangles are known to be Congruent.

Practice Problems

1. Consider triangles △ABC and △ADE,DE is parallel to BC, AD=4cm, DB=6cm, and the base of the triangle BC=12cm. Find the value of DE using Thales Theorem.

2. In △ABC and △DEF, ∠A=∠D, ∠B=∠E, and ∠C=∠F. Are these two triangles similar? Justify your answer.

3. In △PQR and △STU, PQ=6cm, QR=8cm, RP=10cm, ST=3cm, TU=4cm, and US=5cm. Are these two triangles similar? Justify your answer.

4.In △ABC, D and E are points on AB and AC respectively, such that DE ∥ BC. If AD=3cm, DB=6cm, and AC=12cm, find AE.

Related Articles:

• Triangles
• Types of Triangles, Classifications, Practice Questions and FAQs
• Mid Point Theorem