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.
- Definition: Two geometric figures are congruent ($\cong$) if they have the exact same shape and the exact same size.
- Superposition Test: If one figure can be moved (translated, rotated, reflected) and placed perfectly on top of the other figure so that they coincide completely, they are congruent.
- Properties:
- Corresponding angles are equal in measure.
- Corresponding sides are equal in length.
- Example:
- Two new $\textsf{₹}2000$ banknotes (they are designed to be identical in size and shape).
- Two squares with sides measuring 8 cm each.
- Two circles with a radius of 5 cm.

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.
- Definition: Two geometric figures are similar ($\sim$) if they have the same shape but are not necessarily the same size. One figure is an enlargement or a reduction of the other.
- Properties:
- Corresponding angles are equal in measure.
- Corresponding sides are in the same ratio (proportional). This constant ratio is called the scale factor.
- Example:
- A map of a city and the actual city (the map is a reduced, similar version of the city).
- A photograph and a larger print of the same photograph.
- Two squares of different side lengths (e.g., one with 3 cm side, another with 6 cm side). They are both squares (same shape), but different sizes. The ratio of corresponding sides is $3/6 = 1/2$.
- Any two circles are always similar (they have the same shape). The ratio of their radii is the scale factor.
- Two equilateral triangles of different side lengths. They are both equilateral (same shape). The ratio of their side lengths is the scale factor.

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:
- Their corresponding angles are equal in measure.
- 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:
- Vertex A corresponds to Vertex P ($A \leftrightarrow P$)
- Vertex B corresponds to Vertex Q ($B \leftrightarrow Q$)
- Vertex C corresponds to Vertex R ($C \leftrightarrow R$)

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:
- $\angle \text{A} = \angle \text{P}$ (or $\text{m}\angle \text{A} = \text{m}\angle \text{P}$)
- $\angle \text{B} = \angle \text{Q}$ (or $\text{m}\angle \text{B} = \text{m}\angle \text{Q}$)
- $\angle \text{C} = \angle \text{R}$ (or $\text{m}\angle \text{C} = \text{m}\angle \text{R}$)
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 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: 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: 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 $\mathbf{\triangle \text{XYZ} \sim \triangle \text{LMN}}$.
Important Notes on Similarity
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 ($DE \parallel BC$).

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.

(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:
- $\frac{\text{AB}}{\text{DB}} = \frac{\text{AC}}{\text{EC}}$: 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 $\frac{\text{AB}}{\text{DB}} = \frac{\text{AC}}{\text{EC}}$.
- $\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}}$: 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 $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.

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)

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).

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 $\angle \text{C} = \angle \text{R}$.

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:
- $\angle \text{BAC} = \angle \text{QPR}$ (or $\angle \text{A} = \angle \text{P}$) (One pair of equal corresponding angles)
- The sides including these angles are proportional: $\frac{\text{AB}}{\text{PQ}} = \frac{\text{AC}}{\text{PR}}$
Conclusion: $\triangle \text{ABC} \sim \triangle \text{PQR}$ (with correspondence A $\leftrightarrow$ P, B $\leftrightarrow$ Q, C $\leftrightarrow$ R).

Important Comparison between SAS Congruence and SAS Similarity:
- SAS Congruence: Requires two sides and the included angle of one triangle to be equal to the corresponding parts of the other triangle ($\text{Side}_1 = \text{Side'}_1$, $\text{Angle} = \text{Angle'}$, $\text{Side}_2 = \text{Side'}_2$).
- SAS Similarity: Requires the included angle to be equal and the two sides including it to be proportional ($\frac{\text{Side}_1}{\text{Side'}_1} = \frac{\text{Side}_2}{\text{Side'}_2}$, $\text{Angle} = \text{Angle'}$).
Summary of Similarity Criteria
To prove two triangles are similar, check for one of these sets of conditions:
- AA (or AAA): Two (or three) corresponding angles are equal.
- SSS: Corresponding sides are in the same ratio (proportional).
- SAS: Two sides are proportional, and the included angle is equal.
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 $\triangle \text{RPQ} \sim \triangle \text{RTS}$.

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.