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Content On This Page
Congruent Figures vs Similar Figures: Distinction Similarity of Triangles: Definition (Proportional Sides, Equal Angles) Basic Proportionality Theorem (Thales Theorem) and its Converse
Criteria for Similarity of Triangles (AA, SSS, SAS Similarity)


Similarity of Triangles: Concepts and Criteria



Similarity of Triangles: Concepts and Criteria - Congruent Figures vs Similar Figures: Distinction

When comparing geometric figures, we are interested in whether they share properties like shape and size. This leads to two fundamental concepts: congruence and similarity. While related, they describe different levels of correspondence between figures. Understanding the distinction between them is crucial for studying geometric transformations and relationships.


Congruent Figures

As previously discussed in detail, congruent figures are exact replicas of each other.

Two identical squares labelled A and B side-by-side.

In the diagram, Square A and Square B are congruent. If the side length of Square A is $s$, then the side length of Square B is also $s$. Square A $\cong$ Square B.


Similar Figures

Similar figures share the same shape, but not necessarily the same size. One is a scaled version of the other.

Two squares labelled C and D, where D is larger than C.

In the diagram, Square C and Square D are similar. If the side length of Square C is $s_1$ and the side length of Square D is $s_2$, then Square C $\sim$ Square D. The ratio of corresponding sides (e.g., side length of D to side length of C) is the scale factor $k = s_2/s_1$. Here, $s_2 = 2s_1$, so $k=2$.


Key Distinction Summary

Here is a table highlighting the key differences between congruent and similar figures:

Feature Congruent Figures ($\cong$) Similar Figures ($\sim$)
Shape Same Same
Size Same (Ratio of corresponding lengths is 1:1) Can be Different (Ratio of corresponding lengths is a constant scale factor, not necessarily 1)
Corresponding Angles Equal Equal
Corresponding Sides Equal in length Proportional in length (their ratios are equal)
Relationship Congruent figures are a special case of similar figures where the scale factor is 1. Similar figures are not necessarily congruent (unless the scale factor happens to be 1).

In essence, congruence is a more restrictive condition than similarity. If figures are congruent, they are always similar. But if they are similar, they are only congruent if their size happens to be the same (scale factor is 1).



Similarity of Triangles: Concepts and Criteria - Similarity of Triangles: Definition (Proportional Sides, Equal Angles)

Expanding on the general concept of similar figures, we now focus on the definition of similar triangles. Triangles are similar if they have the same shape, even if they differ in size. This geometric relationship is fundamental to many concepts in geometry and trigonometry, including scale drawings and the properties of parallel lines and transversals.


Definition of Similar Triangles

Two triangles are defined as similar if and only if they satisfy the following two conditions simultaneously regarding their corresponding parts:

  1. Their corresponding angles are equal in measure.
  2. Their corresponding sides are in the same ratio (proportional).

If $\triangle \text{ABC}$ is similar to $\triangle \text{PQR}$, we denote this relationship using the similarity symbol ($\sim$) as:

$\triangle \text{ABC} \sim \triangle \text{PQR}$

Just like with congruence, the order of the vertices in the similarity statement is critically important. It indicates the precise correspondence between the vertices of the two triangles:

Two triangles ABC and PQR. Triangle PQR is larger than ABC. Corresponding angles (A=P, B=Q, C=R) are marked equal. Corresponding sides are labelled.

This similarity statement $\triangle \text{ABC} \sim \triangle \text{PQR}$ implies the following equalities between corresponding angles and proportionalities between corresponding sides:

1. Equality of Corresponding Angles:

The angles at corresponding vertices are equal in measure:

2. Proportionality of Corresponding Sides:

The lengths of the sides opposite to corresponding angles are in the same constant ratio. This ratio is called the scale factor or the ratio of similarity.

The side opposite $\angle \text{A}$ (BC) corresponds to the side opposite $\angle \text{P}$ (QR). The side opposite $\angle \text{B}$ (AC) corresponds to the side opposite $\angle \text{Q}$ (PR). The side opposite $\angle \text{C}$ (AB) corresponds to the side opposite $\angle \text{R}$ (PQ).

So, the ratios of the lengths of corresponding sides are equal:

$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}} = k$

(where $k$ is the scale factor)

The scale factor $k$ represents the ratio by which one triangle is scaled to obtain the other. If we write the ratio as side in $\triangle \text{ABC}$ / side in $\triangle \text{PQR}$, then $k$ is the factor by which $\triangle \text{PQR}$ needs to be scaled to get $\triangle \text{ABC}$. If we write it as side in $\triangle \text{PQR}$ / side in $\triangle \text{ABC}$, the scale factor is $1/k$. The choice of which triangle is the "original" and which is the "scaled" version determines whether $k>1$ (enlargement) or $0


Important Notes on Similarity

It is a significant property of triangles that the two conditions for similarity are dependent. If the corresponding angles of two triangles are equal, then their corresponding sides must be proportional, and vice-versa. This fact leads to the development of specific criteria (postulates/theorems) for proving triangle similarity by checking only a subset of these conditions, similar to congruence criteria.


Example 1. $\triangle \text{XYZ}$ has angles $\text{m}\angle \text{X} = 40^\circ$, $\text{m}\angle \text{Y} = 60^\circ$. $\triangle \text{LMN}$ has angles $\text{m}\angle \text{L} = 40^\circ$, $\text{m}\angle \text{N} = 80^\circ$. Are these triangles similar? If yes, state the correspondence.

Answer:

Given:

  • In $\triangle \text{XYZ}$: $\text{m}\angle \text{X} = 40^\circ$, $\text{m}\angle \text{Y} = 60^\circ$.
  • In $\triangle \text{LMN}$: $\text{m}\angle \text{L} = 40^\circ$, $\text{m}\angle \text{N} = 80^\circ$.

To Determine: Are the triangles similar? If yes, state the correspondence.

Solution:

To check for similarity based on angles, we need to find all three angles in each triangle and compare corresponding angles.

In $\triangle \text{XYZ}$, using the Angle Sum Property:

$\text{m}\angle \text{X} + \text{m}\angle \text{Y} + \text{m}\angle \text{Z} = 180^\circ$

$40^\circ + 60^\circ + \text{m}\angle \text{Z} = 180^\circ$

$100^\circ + \text{m}\angle \text{Z} = 180^\circ$

$\text{m}\angle \text{Z} = 180^\circ - 100^\circ = 80^\circ$

The angles of $\triangle \text{XYZ}$ are $40^\circ, 60^\circ, 80^\circ$.

In $\triangle \text{LMN}$, using the Angle Sum Property:

$\text{m}\angle \text{L} + \text{m}\angle \text{M} + \text{m}\angle \text{N} = 180^\circ$

$40^\circ + \text{m}\angle \text{M} + 80^\circ = 180^\circ$

$120^\circ + \text{m}\angle \text{M} = 180^\circ$

$\text{m}\angle \text{M} = 180^\circ - 120^\circ = 60^\circ$

The angles of $\triangle \text{LMN}$ are $40^\circ, 60^\circ, 80^\circ$.

Comparing the angles, we see that the angle measures are the same in both triangles: $\{40^\circ, 60^\circ, 80^\circ\}$.

We can identify the corresponding angles:

  • $\angle \text{X} = 40^\circ$, $\angle \text{L} = 40^\circ \implies \angle \text{X}$ corresponds to $\angle \text{L}$.
  • $\angle \text{Y} = 60^\circ$, $\angle \text{M} = 60^\circ \implies \angle \text{Y}$ corresponds to $\angle \text{M}$.
  • $\angle \text{Z} = 80^\circ$, $\angle \text{N} = 80^\circ \implies \angle \text{Z}$ corresponds to $\angle \text{N}$.

Since all corresponding angles are equal, the triangles are similar by the definition of similar triangles (and the AA similarity criterion, which we will cover, states that this is sufficient).

The correspondence between vertices is X $\leftrightarrow$ L, Y $\leftrightarrow$ M, Z $\leftrightarrow$ N.

Conclusion: Yes, $\triangle \text{XYZ}$ and $\triangle \text{LMN}$ are similar. The similarity is written as $\mathbf{\triangle \text{XYZ} \sim \triangle \text{LMN}}$.



Similarity of Triangles: Concepts and Criteria - Basic Proportionality Theorem (Thales Theorem) and its Converse

The Basic Proportionality Theorem (BPT), also famously known as Thales Theorem, is a cornerstone theorem in the study of similar triangles and proportional segments within a triangle. It provides a specific condition involving a line parallel to one side of a triangle and its intersection with the other two sides.


Basic Proportionality Theorem (BPT) / Thales Theorem

Theorem Statement: If a line is drawn parallel to one side of a triangle so as to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Given: In $\triangle \text{ABC}$, a line DE is drawn such that it intersects side AB at point D and side AC at point E, and DE is parallel to side BC ($DE \parallel BC$).

Triangle ABC with line DE parallel to BC, intersecting AB at D and AC at E.

To Prove: $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.

Construction: Join the points B to E and C to D. Draw a line segment EM perpendicular to AB and a line segment DN perpendicular to AC.

Triangle ABC with line DE parallel to BC. Construction lines BE and CD are drawn. Perpendiculars EM to AB and DN to AC are drawn.

(Here, EM is the height of $\triangle \text{ADE}$ and $\triangle \text{BDE}$ with respect to bases AD and DB respectively. DN is the height of $\triangle \text{ADE}$ and $\triangle \text{CDE}$ with respect to bases AE and EC respectively).

Proof:

We know that the area of a triangle is given by the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

Consider $\triangle \text{ADE}$ and $\triangle \text{BDE}$ sharing the same height EM.

Area($\triangle \text{ADE}$) $= \frac{1}{2} \times \text{AD} \times \text{EM}$

Area($\triangle \text{BDE}$) $= \frac{1}{2} \times \text{DB} \times \text{EM}$

Taking the ratio of their areas:

$\frac{\text{Area}(\triangle \text{ADE})}{\text{Area}(\triangle \text{BDE})} = \frac{\frac{1}{2} \times \text{AD} \times \text{EM}}{\frac{1}{2} \times \text{DB} \times \text{EM}}$

$\frac{\text{Area}(\triangle \text{ADE})}{\text{Area}(\triangle \text{BDE})} = \frac{\text{AD}}{\text{DB}}$

... (i)

Now, consider $\triangle \text{ADE}$ and $\triangle \text{CDE}$ sharing the same height DN.

Area($\triangle \text{ADE}$) $= \frac{1}{2} \times \text{AE} \times \text{DN}$

Area($\triangle \text{CDE}$) $= \frac{1}{2} \times \text{EC} \times \text{DN}$

Taking the ratio of their areas:

$\frac{\text{Area}(\triangle \text{ADE})}{\text{Area}(\triangle \text{CDE})} = \frac{\frac{1}{2} \times \text{AE} \times \text{DN}}{\frac{1}{2} \times \text{EC} \times \text{DN}}$

$\frac{\text{Area}(\triangle \text{ADE})}{\text{Area}(\triangle \text{CDE})} = \frac{\text{AE}}{\text{EC}}$

... (ii)

Now, consider $\triangle \text{BDE}$ and $\triangle \text{CDE}$. Both triangles lie on the same base DE and are between the same parallel lines DE and BC (since $DE \parallel BC$ is given).

$\text{Area}(\triangle \text{BDE}) = \text{Area}(\triangle \text{CDE})$

(Triangles on the same base and between the same parallels are equal in area)

Let's call this equation (iii).

... (iii)

From equations (i), (ii), and (iii):

Substitute $\text{Area}(\triangle \text{CDE})$ with $\text{Area}(\triangle \text{BDE})$ in equation (ii):

$\frac{\text{Area}(\triangle \text{ADE})}{\text{Area}(\triangle \text{BDE})} = \frac{\text{AE}}{\text{EC}}$

(Using iii in ii)

Now, compare this modified equation (ii) with equation (i). Both ratios have the same numerator ($\text{Area}(\triangle \text{ADE})$) and equal denominators ($\text{Area}(\triangle \text{BDE})$). Thus, the ratios must be equal.

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

(From i and modified ii)

Hence, the Basic Proportionality Theorem is proved: the line DE divides the other two sides AB and AC in the same ratio.

Hence Proved.

Corollaries of BPT:

From the basic proportionality $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$, we can derive other useful proportionalities:

  1. $\frac{\text{AB}}{\text{DB}} = \frac{\text{AC}}{\text{EC}}$: Add 1 to both sides of $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$: $\frac{\text{AD}}{\text{DB}} + 1 = \frac{\text{AE}}{\text{EC}} + 1 \implies \frac{\text{AD} + \text{DB}}{\text{DB}} = \frac{\text{AE} + \text{EC}}{\text{EC}}$. Since $\text{AD} + \text{DB} = \text{AB}$ and $\text{AE} + \text{EC} = \text{AC}$, we get $\frac{\text{AB}}{\text{DB}} = \frac{\text{AC}}{\text{EC}}$.
  2. $\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}}$: Take the reciprocal of $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$: $\frac{\text{DB}}{\text{AD}} = \frac{\text{EC}}{\text{AE}}$. Add 1 to both sides: $\frac{\text{DB}}{\text{AD}} + 1 = \frac{\text{EC}}{\text{AE}} + 1 \implies \frac{\text{DB} + \text{AD}}{\text{AD}} = \frac{\text{EC} + \text{AE}}{\text{AE}} \implies \frac{\text{AB}}{\text{AD}} = \frac{\text{AC}}{\text{AE}}$. Now take the reciprocal again: $\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}}$.

These corollaries provide alternative ways to express the proportionality of sides when a line parallel to one side intersects the other two.


Converse of the Basic Proportionality Theorem

The converse of the BPT is also a significant theorem. It provides a condition for determining if a line segment within a triangle is parallel to one of its sides.

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

Given: In $\triangle \text{ABC}$, a line DE intersects side AB at point D and side AC at point E, such that $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.

Triangle ABC with line DE intersecting AB at D and AC at E, with the ratio AD/DB = AE/EC.

To Prove: $DE \parallel BC$.

Proof:

We will prove this theorem using the method of contradiction.

Assume that DE is not parallel to BC.

(Assumption for contradiction)

If DE is not parallel to BC, then there must exist another line passing through point D which is parallel to BC (by Playfair's Axiom/Euclid's Fifth Postulate).

Let's draw a line DF through D such that DF is parallel to BC, where F is a point on AC ($F \neq E$).

(Construction for contradiction)

Triangle ABC with line DE intersecting AB and AC. A line DF is drawn parallel to BC, where F is on AC and shown as a distinct point from E.

Since DF || BC and DF intersects AB and AC in distinct points D and F, by the Basic Proportionality Theorem:

$\frac{\text{AD}}{\text{DB}} = \frac{\text{AF}}{\text{FC}}$

(By BPT, since DF || BC) ... (i)

However, we are given that the line DE divides AB and AC in the same ratio:

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

(Given) ... (ii)

From equations (i) and (ii), since both ratios are equal to $\frac{\text{AD}}{\text{DB}}$, we can equate the other parts:

$\frac{\text{AF}}{\text{FC}} = \frac{\text{AE}}{\text{EC}}$

(From i and ii)

Add 1 to both sides of this equation:

$\frac{\text{AF}}{\text{FC}} + 1 = \frac{\text{AE}}{\text{EC}} + 1$

(Adding equals to equals - Common Notion 2)

$\frac{\text{AF} + \text{FC}}{\text{FC}} = \frac{\text{AE} + \text{EC}}{\text{EC}}$

(Simplifying the fractions)

From the figure, AF + FC is the entire length of AC, and AE + EC is also the entire length of AC (assuming F and E are on AC).

$\text{AF} + \text{FC} = \text{AC}$

(Segment Addition Postulate)

$\text{AE} + \text{EC} = \text{AC}$

(Segment Addition Postulate)

Substitute AC into the ratio equation:

$\frac{\text{AC}}{\text{FC}} = \frac{\text{AC}}{\text{EC}}$

(Substitution)

Since AC is a side length, $\text{AC} \neq 0$. We can effectively "cancel" AC from both sides or take reciprocals and then multiply by AC.

$\frac{1}{\text{FC}} = \frac{1}{\text{EC}}$

(Dividing both sides by AC)

$\text{FC} = \text{EC}$

(Taking reciprocals)

This equality, FC = EC, means that point F and point E must be the same point on the line AC.

(This contradicts our assumption that $F \neq E$)

Therefore, the line DF (which we constructed to be parallel to BC) must coincide with the line DE.

(Since F coincides with E)

Since DF is parallel to BC, it follows that DE must be parallel to BC.

(As DE is the same line as DF)

Our initial assumption that DE is not parallel to BC led to a contradiction (that E and F are distinct points, but then we showed they must be the same point). Therefore, the assumption must be false, and DE must be parallel to BC.

Hence, the converse of the Basic Proportionality Theorem is proved.

Hence Proved.


Example 1. In $\triangle \text{PQR}$, points S and T are on sides PQ and PR respectively. If PS = 3 cm, SQ = 9 cm, PT = 4 cm, and TR = 12 cm, show that ST || QR.

Answer:

Given: In $\triangle \text{PQR}$, S is on PQ and T is on PR. PS = 3 cm, SQ = 9 cm, PT = 4 cm, TR = 12 cm.

To Prove: ST || QR.

Solution:

To show that ST is parallel to QR, we can use the converse of the Basic Proportionality Theorem. We need to check if the line segment ST divides the sides PQ and PR in the same ratio.

Calculate the ratio of the segments on side PQ:

$\frac{\text{PS}}{\text{SQ}} = \frac{3 \text{ cm}}{9 \text{ cm}}$

$= \frac{1}{3}$

Calculate the ratio of the segments on side PR:

$\frac{\text{PT}}{\text{TR}} = \frac{4 \text{ cm}}{12 \text{ cm}}$

$= \frac{1}{3}$

Comparing the two ratios, we see that:

$\frac{\text{PS}}{\text{SQ}} = \frac{\text{PT}}{\text{TR}}$

(Both ratios are equal to 1/3)

Since the line segment ST divides the sides PQ and PR in the same ratio, by the converse of the Basic Proportionality Theorem, the line segment ST must be parallel to the third side QR.

Therefore, ST || QR.

Hence, shown.



Similarity of Triangles: Concepts and Criteria - Criteria for Similarity of Triangles (AA, SSS, SAS Similarity)

The definition of similar triangles requires that corresponding angles are equal AND corresponding sides are proportional. However, just like with congruence, we have simpler criteria that allow us to prove similarity by checking only a minimum subset of these conditions. These criteria are theorems that are derived from the definition of similarity and the properties of triangles (like the Angle Sum Property and Basic Proportionality Theorem).


Minimum Conditions for Triangle Similarity

If any one of the following criteria is satisfied by two triangles, it is sufficient to conclude that they are similar.

1. AA (Angle-Angle) Similarity Criterion

Statement: If two angles of one triangle are respectively equal to two corresponding angles of another triangle, then the two triangles are similar.

Reasoning: This is the most commonly used criterion. If two pairs of corresponding angles are equal, the third pair of angles must also be equal because the sum of the angles in any triangle is $180^\circ$. If $\angle \text{A} = \angle \text{P}$ and $\angle \text{B} = \angle \text{Q}$, then $\angle \text{C} = 180^\circ - (\angle \text{A} + \angle \text{B})$ and $\angle \text{R} = 180^\circ - (\angle \text{P} + \angle \text{Q})$. Since $\angle \text{A} = \angle \text{P}$ and $\angle \text{B} = \angle \text{Q}$, it follows that $\angle \text{C} = \angle \text{R}$. Thus, all three pairs of corresponding angles are equal, which is one of the defining properties of similar triangles. (The proportionality of sides then follows from this).

Condition: In $\triangle \text{ABC}$ and $\triangle \text{PQR}$, if $\angle \text{A} = \angle \text{P}$ and $\angle \text{B} = \angle \text{Q}$.

Conclusion: $\triangle \text{ABC} \sim \triangle \text{PQR}$ (with correspondence A $\leftrightarrow$ P, B $\leftrightarrow$ Q, C $\leftrightarrow$ R).

Two triangles ABC and PQR. Angle A is marked equal to Angle P, and Angle B is marked equal to Angle Q.

This criterion is also sometimes stated as AAA (Angle-Angle-Angle) similarity, but AA is sufficient because the third angle equality is automatic.


2. SSS (Side-Side-Side) Similarity Criterion

Statement: If the corresponding sides of two triangles are in the same ratio (proportional), then their corresponding angles are equal, and hence the triangles are similar.

Condition: In $\triangle \text{ABC}$ and $\triangle \text{PQR}$, if the ratio of corresponding sides is constant:

$\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}}$

Conclusion: $\triangle \text{ABC} \sim \triangle \text{PQR}$ (with correspondence A $\leftrightarrow$ P, B $\leftrightarrow$ Q, C $\leftrightarrow$ R). This conclusion also implies that $\angle \text{A} = \angle \text{P}$, $\angle \text{B} = \angle \text{Q}$, and $\angle \text{C} = \angle \text{R}$.

Two triangles ABC and PQR where the ratio of corresponding sides AB/PQ = BC/QR = AC/PR is constant, indicated by side labels like 2x, 2y, 2z and x, y, z.

3. SAS (Side-Angle-Side) Similarity Criterion

Statement: If one angle of a triangle is equal to one corresponding angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

Condition: In $\triangle \text{ABC}$ and $\triangle \text{PQR}$, if:

Conclusion: $\triangle \text{ABC} \sim \triangle \text{PQR}$ (with correspondence A $\leftrightarrow$ P, B $\leftrightarrow$ Q, C $\leftrightarrow$ R).

Two triangles ABC and PQR. Angle A is marked equal to Angle P. Sides AB and AC in ABC and PQ and PR in PQR are indicated to be proportional.

Important Comparison between SAS Congruence and SAS Similarity:


Summary of Similarity Criteria

To prove two triangles are similar, check for one of these sets of conditions:

Note that there are no SSA or ASA/AAS similarity criteria analogous to congruence. AA similarity effectively covers ASA/AAS because knowing two angles determines the third.


Example 1. In the figure, $\angle \text{P} = \angle \text{RTS}$. Prove that $\triangle \text{RPQ} \sim \triangle \text{RTS}$.

Triangle RPQ with point S on RP and point T on RQ such that ST is joined, forming triangle RTS. Angle P is marked equal to Angle RTS.

Answer:

Given: In $\triangle \text{RPQ}$, points S and T are on sides RP and RQ respectively. $\angle \text{P} = \angle \text{RTS}$.

To Prove: $\triangle \text{RPQ} \sim \triangle \text{RTS}$.

Proof:

Consider the two triangles $\triangle \text{RPQ}$ and $\triangle \text{RTS}$.

1. $\angle \text{RPQ} = \angle \text{RTS}$

(Given)

2. $\angle \text{PRQ} = \angle \text{SRT}$

(Common angle to both triangles, which is $\angle \text{R}$)

We have shown that two angles of $\triangle \text{RPQ}$ are equal to two corresponding angles of $\triangle \text{RTS}$.

Therefore, by the AA (Angle-Angle) similarity criterion:

$\triangle \text{RPQ} \sim \triangle \text{RTS}$

(AA Similarity Criterion)

Note the order of vertices in the similarity statement matches the corresponding angles: R $\leftrightarrow$ R (common angle), P $\leftrightarrow$ T (given equal angles), Q $\leftrightarrow$ S (the remaining third angles must be equal).

Hence, proved.