Criteria for Similarity of Triangles

A triangle is the simplest polygon in mathematics. It is a closed two-dimensional figure formed by three sides and three angles. The sum of the interior angles of a triangle is always 180°.

Similar Triangles: Two triangles are called similar if they have the same shape but may differ in size. In similar triangles, the corresponding angles are equal, and the corresponding sides are proportional.

The figure above shows two triangles, ΔABC and ΔPQR, having the same shape but different sizes. Their corresponding angles are equal, and sides are proportional, so the triangles are similar.

Conditions for Similarity of Two Triangles

Consider two triangles, ABC and DEF.

Two triangles are said to be similar if:

• Their corresponding angles are equal
  ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
• Their corresponding sides are proportional
  AB/DE = BC/EF = CA/FD

In this case, the triangles are similar.
Therefore, ΔABC ∼ ΔDEF.

The similarity of two triangles can be verified using the following criteria:

1. SSS Similarity Criterion for Two Triangles

It states that 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.

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

Proof:

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

2. AAA Similarity Criterion for Two Triangles

AAA refers to angles (all three) of the triangles. It states, “If the two corresponding angles of both triangles are equal, then their respective sides will always have the 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 triangle equal, as the sum is always 180°.

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

Proof:

Given:
∠A = ∠D
∠B = ∠E
∠C = ∠F

In triangles ABC and DEF, two angles are equal, so the third angle is also equal (sum of angles in a triangle is 180°).

Therefore,
ΔABC ∼ ΔDEF (by AA similarity criterion)

Hence, corresponding sides are proportional:

AB/DE = AC/DF = BC/EF

3. SAS Similarity Criterion for Two Triangles

If two sides of one triangle are proportional to the corresponding two sides of another triangle, and the included angle between those sides is equal, then the triangles are similar.

To Prove: ΔABC∼ ΔDEF

Proof:

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

Related Articles:

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

Sample Problems

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

AC² = AB² + BC²

p² = (4q)² + (3q)²
p² = 16q² + 9q²
p² = 25q²
p = 5q

Question 3: In the figure given below, 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 2 cm. 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 are the SAS and SSS criteria 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's 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.