Similar triangles are triangles with the same shape but can have variable sizes. Having corresponding sides in proportion to each other and corresponding angles equal to each other. Similar triangles are different from congruent triangles. Two congruent figures are always similar, but two similar figures need not be congruent.
Two triangles are considered similar when their corresponding angles match, and their sides are proportional. This means that similar triangles have the same shape, although their sizes may differ. On the other hand, triangles are defined as congruent when they not only share the same shape but also have corresponding sides that are identical in length.
When two triangles are similar, it implies that:
- All pairs of corresponding angles in the triangles are equal.
- All pairs of corresponding sides of the triangle are proportional.
The symbol "∼" is used to represent the similarity between similar triangles. So, when two triangles are similar, we write it as △ABC ∼ △DEF.
Examples
- If we take two triangles that have sides in the same ratio, then they are similar triangles.
- The flagpoles and their shadows represent similar triangles.
The triangles shown in the image below are similar, and we represent them as △ABC ∼ △PQR.
Basic Proportionality Theorem (Thales' Theorem)
The Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides in the same ratio.
If in triangle ABC, a line DE is drawn parallel to BC intersecting AB at D and AC at E, then:
AD/DB = AE/EC
Since DE ∥ BC, corresponding angles are equal, making triangles ADE and ABC similar. Hence, their corresponding sides are proportional.
Similar Triangle Rules
The similarity theorems help us to find whether the two triangles are similar or not. When we do not have the measure of angles or the sides of the triangles, we use the similarity theorems.
There are three major types of similarity rules, as given below:
- AA (or AAA) or Angle-Angle Similarity Theorem
- SAS or Side-Angle-Side Similarity Theorem
- SSS or Side-Side-Side Similarity Theorem
Angle-Angle (AA) or AAA Similarity Theorem
The AA similarity criterion states that if any two angles in a triangle are respectively equal to any two angles of another triangle, then they must be similar triangles. The AA similarity rule is easily applied when we only know the measure of the angles and have no idea about the length of the sides of the triangle.
In the image given below, if it is known that ∠B = ∠G and ∠C = ∠F:
And we can say that by the AA similarity criterion, △ABC and △EGF are similar, or △ABC ∼ △EGF.
⇒AB/EG = BC/GF = AC/EF and ∠A = ∠E.
Side-Angle-Side, or SAS, Similarity Theorem
According to the SAS similarity theorem, if any two sides of the first triangle are in exact proportion to the two sides of the second triangle, along with the angle formed by these two sides of the individual triangles being equal, then they must be similar triangles. This rule is generally applied when we only know the measure of two sides and the angle formed between those two sides in both triangles, respectively.
In the image given below, if it is known that AB/DE = AC/DF and ∠A = ∠D
And we can say that by the SAS similarity criterion, △ABC and △DEF are similar or △ABC ∼ △DEF.
Side-Side-Side, or SSS, Similarity Theorem
According to the SSS similarity theorem, two triangles will be similar to each other if the corresponding ratios of all the sides of the two triangles are equal. This criterion is commonly used when we only have the measure of the sides of the triangle and have less information about the angles of the triangle.
In the image given below, if it is known that PQ/ED = PR/EF = QR/DF
And we can say that by the SSS similarity criterion, △PQR and △EDF are similar, or △PQR ∼ △EDF.
Similar Triangles Properties
Similar triangles have various properties that are widely used for solving various geometrical problems. Some of the common properties of similar triangles:
- The shape of similar triangles is fixed, but their sizes may be different.
- Corresponding angles of similar triangles are equal.
- Corresponding sides of similar triangles are in common ratios.
- The ratio of the area of similar triangles is equal to the square of the ratio of their corresponding sides.
How to Find Similar Triangles?
Two given triangles can be proved as similar triangles using the above-given theorems. We can follow the steps given below to check if the given triangles are similar or not:
Step 1: Note down the given dimensions of the triangles (corresponding sides or corresponding angles).
Step 2: Check if these dimensions follow any of the conditions for similar triangles theorems (AA, SSS, SAS).
Step 3: The given triangles, if they satisfy any of the similarity theorems, can be represented using the "∼" to denote similarity.
This can be understood better with the help of the following example:
Example: Check if △ABC and △PQR are similar triangles or not using the given data: ∠A = 65°, ∠B = 70°, ∠P = 70°, and ∠R = 45°.
Using the given angles, first find the third angle in each triangle using the angle sum property.
We know that in a triangle, the sum of all angles is 180°.In △ABC, ∠C = 180° − (∠A + ∠B) = 180° − (65° + 70°) = 45°
In △PQR, ∠Q = 180° − (∠P + ∠R) = 180° − (70° + 45°) = 65°
Now comparing the angles,
∠A = ∠Q = 65°, ∠B = ∠P = 70°, and ∠C = ∠R = 45°Since the corresponding angles are equal, △ABC and △PQR are similar by AA similarity criterion.
Therefore, △ABC ∼ △QPR.
Area of Similar Triangles - Theorem
The Similar Triangle Area Theorem states that for two similar triangles, the ratio of the area of the triangles is proportional to the square of the ratio of their corresponding sides. Suppose we are given two similar triangles, ΔABC and ΔPQR; then
According to the Similar Triangle Theorem:
(Area of ΔABC)/(Area of ΔPQR) = (AB/PQ)2 = (BC/QR)2 = (CA/RP)2
Applications of Similar Triangles
Various applications of the similar triangle that we see in real life are,
- The shadow and height of various objects are calculated using the concept of similar triangles.
- Map scaling uses the concept of a similar triangle.
- Photographic devices use similar triangle properties to capture various images.
- Model making uses the concept of similar triangles.
- Navigation and Trigonometry also use a similar triangle approach to solve various problems, etc.
- The ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides.
- All congruent triangles are similar, but all similar triangles may not necessarily be congruent.
- This ‘~’ symbol is used to denote similar triangles.
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Solved Questions
Question 1: In Figure 1, DE || BC. If AD = 1.7 cm, AB = 6.8 cm, and AC = 9 cm. Find AE?
Solution:
Let AE = x cm.
In △ABC, DE || BC
By Thales Theorem we have,
AD/AB = AE/AC
1.7/6.8 = x/9
x = (1.7×9)/6.8 = 2.25 cm
AE = 2.25 cm
Hence AE = 2.25 cm
Question 2: In the given figure 1, DE || BC. If AD = 2.5 cm, DB = 3 cm, and AE = 3.75 cm. Find AC?
Solution:
In △ABC, DE || BC
AD/DB = AE/EC (By Thales' Theorem)
2.5/3 = 3.75/x, where EC = x cm
(3 × 3.75)/2.5 = 9/2 = 4.5 cm
EC = 4.5 cm
Hence, AC = (AE + EC) = 3.75 + 4.5 = 8.25 cm.
Question 3: If D and E are points on the sides AB and AC, respectively, of △ABC (figure 2) such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm, show that DE || || BC.
Solution:
Given, AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
AD/AB = 1.4/5.6 = 1/4 and AE/AC = 1.8/7.2 = 1/4
AD/AB = AE/AC
Hence, by converse of Thales Theorem, DE || BC.
Question 4: Prove that the line segment joining the midpoints of any two sides of a triangle (figure 2) is parallel to the third side.
Solution:
In △ABC in which D and E are the midpoints of AB and AC respectively.
Since D and E are the midpoints of AB and AC respectively, we have :
AD = DB and AE = EC.
AD/DB = AE/EC (each equal to 1)
Hence, by converse of Thales Theorem, DE || BC
Question 5: Prove that a line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.
Solution:
Given a ΔΑΒC in which D is the midpoint of AB and DE || BC, meeting AC at E.
TO PROVE AE = EC.
Proof: Since DE || BC, by Thales' theorem, we have:
AE/AD = EC/DB =1 (AD = DB, given)
AE/EC = 1
AE = EC
Question 6: In the given Figure 2, AD/DB = AE/EC and ∠ADE = ∠ACB. Prove that ABC is an isosceles triangle.
Solution:
We have AD/DB = AE/EC DE || BC [by the converse of Thales' theorem]
∠ADE = ∠ABC (corresponding ∠s)
But, ∠ADE = ∠ACB (given).
Hence, ∠ABC = ∠ACB.
So, AB = AC [sides opposite to equal angles].
Hence, △ABC is an isosceles triangle.
Practice Questions
Q1. In two similar triangles, △ABC and △ADE, if DE || || BC and AD = 3 cm, AB = 8 cm, and AC = 6 cm. Find AE.
Q2. In two similar triangles, △ABC and △PQR, if QR = || BC and PQ = 2 cm, AB = 12 cm, and AC = 9 cm. Find PR.
Q3. In two similar triangles, ΔABC and ΔAPQ, the lengths of the sides are given as AP = 3 cm, PB = 4 cm, and BC = 8 cm. Find the ratio of the areas of ΔABC and ΔAPQ.
Q4. In two similar triangles, ΔABC and ΔAPQ, the lengths of the sides are given as AP = 9 cm, PB = 12 cm, and BC = 24 cm. Find the ratio of the areas of ΔABC and ΔAPQ.