Class 10 Mathematics: Triangles – Basic Proportionality Theorem & Similarity Criteria

In geometry, shape and size define how figures interact with one another. While two-dimensional objects with the exact same shape and exact same size are termed congruent, objects that share the exact same shape regardless of their size are called similar.

The concept of similarity forms the foundation of Euclidean geometry, trigonometry, coordinate geometry, and real-life spatial scaling. In the Class 10 NCERT/CBSE curriculum, the study of similar triangles focuses on two central pillars:

1. The Basic Proportionality Theorem (BPT), historically known as Thales Theorem.
2. Criteria for Similarity of Triangles ($AAA$, $SSS$, and $SAS$).

Understanding these core concepts allows us to calculate unmeasurable distances, prove geometric proportions, and analyze complex figures by breaking them down into manageable, proportional relationships.

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In-Depth Conceptual Breakdown

1. Geometric Similarity vs. Congruence

Two geometric figures are similar if they have the same shape, even if one is a scaled-up or scaled-down version of the other. For two polygons with the same number of sides to be similar, two conditions must be satisfied simultaneously:

1. All corresponding angles are equal.
2. All corresponding sides are in the same ratio (or proportion).

   Congruent Triangles (Same Size, Same Shape)
   △ABC ≅ △DEF  ⇒  AB = DE, BC = EF, AC = DF
                   ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

   Similar Triangles (Different Size, Same Shape)
   △ABC ~ △DEF  ⇒  AB/DE = BC/EF = AC/DF
                   ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

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2. Basic Proportionality Theorem (Thales Theorem)

The Basic Proportionality Theorem establishes a fundamental relationship between lines drawn parallel to one side of a triangle and the segments created on the remaining two sides.

Theorem Statement

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

Formal Proof (Standard Board Pattern)

                  A
                 / \
                /   \
               /  D  \
              D-------E
             /  .   .  \
            /     .     \
           /        .    \
          B---------------C

• Given: A triangle $ABC$ in which a line parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$. Thus, $DE \parallel BC$.

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

• Construction: Join $BE$ and $CD$. Draw $EF \perp AB$ and $DG \perp AC$.

• Proof: The area of a triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.

  In $\triangle ADE$, taking $AD$ as the base, the height is $EF$: $\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EF \quad \text{--- (1)}$

  In $\triangle BDE$, taking $DB$ as the base, the height is also $EF$ (since $\triangle BDE$ is an obtuse triangle relative to base $DB$): $\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EF \quad \text{--- (2)}$

  Dividing equation (1) by equation (2): $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB} \quad \text{--- (3)}$

  Similarly, considering $\triangle ADE$ with base $AE$ and altitude $DG$, and $\triangle DEC$ with base $EC$ and altitude $DG$: $\text{Area}(\triangle ADE) = \frac{1}{2} \times AE \times DG$ $\text{Area}(\triangle DEC) = \frac{1}{2} \times EC \times DG$

  Dividing these two areas: $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC} \quad \text{--- (4)}$

  Notice that $\triangle BDE$ and $\triangle DEC$ lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$. By standard geometric theorems: $\text{Area}(\triangle BDE) = \text{Area}(\triangle DEC) \quad \text{--- (5)}$

  Substituting equation (5) into equation (4), we get: $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{AE}{EC} \quad \text{--- (6)}$

  Equating the left-hand sides of equations (3) and (6): $\frac{AD}{DB} = \frac{AE}{EC}$

  (Hence Proved)

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Important Corollaries of BPT

By manipulating the primary relation $\frac{AD}{DB} = \frac{AE}{EC}$, we can derive two additional useful forms:

1. Adding $1$ to both sides: $\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1 \implies \frac{AD + DB}{DB} = \frac{AE + EC}{EC} \implies \frac{AB}{DB} = \frac{AC}{EC}$

2. Inverting and adding $1$ to both sides: $\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1 \implies \frac{DB + AD}{AD} = \frac{EC + AE}{AE} \implies \frac{AB}{AD} = \frac{AC}{AE}$

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3. Converse of the Basic Proportionality Theorem

Theorem Statement

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

• Mathematical Statement: In $\triangle ABC$, if points $D$ and $E$ lie on $AB$ and $AC$ respectively such that: $\frac{AD}{DB} = \frac{AE}{EC}$ then $DE \parallel BC$.

• Method of Proof: Proof by Contradiction. If $DE$ is assumed not parallel to $BC$, we draw a line $DE'$ parallel to $BC$. Applying BPT to $DE'$ yields $\frac{AD}{DB} = \frac{AE'}{E'C}$. Equating this to the given condition shows $E$ and $E'$ must coincide, proving $DE \parallel BC$.

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4. Criteria for Similarity of Triangles

To prove that two triangles are similar, we do not need to measure all three angles and all three sides. Certain minimal combinations of equal angles or proportional sides are sufficient.

                   A                                D
                  / \                              / \
                 /   \                            /   \
                /     \                          /     \
               B-------C                        E-------F

Criteria Summary Table

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Detailed Breakdown of Criteria

1. AAA (Angle-Angle-Angle) & AA Similarity

If two triangles have their corresponding angles equal, their corresponding sides are automatically proportional.

• AA Corollary: Because the sum of angles in any triangle is always $180^\circ$, if two angles of one triangle equal two angles of another ($\angle A = \angle D$ and $\angle B = \angle E$), the third angles must also be equal ($\angle C = 180^\circ - (\angle A + \angle B) = \angle F$). Therefore, AA similarity is the most commonly used form in exam proofs.

2. SSS (Side-Side-Side) Similarity

If the corresponding sides of two triangles are in the same ratio, their corresponding angles are automatically equal, making the triangles similar. $\text{If } \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = k, \text{ then } \triangle ABC \sim \triangle DEF$

3. SAS (Side-Angle-Side) Similarity

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

• Crucial Rule: The angle must be enclosed directly between the two proportional sides.
  • Correct: $\frac{AB}{DE} = \frac{AC}{DF}$ with included angle $\angle A = \angle D$.
  • Incorrect: $\frac{AB}{DE} = \frac{AC}{DF}$ with non-included angle $\angle B = \angle E$ (does not guarantee similarity).

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Real-World Applications

1. Indirect Height Measurement (Shadow & Reflection Method)

Before modern laser measuring devices, surveyors, civil engineers, and astronomers measured the heights of inaccessible objects (trees, pyramids, towers) using triangle similarity.

       Sun Rays
        \
         \   | Tree (Height H)
          \  |
           \ |________ Shadow S1
             \
              \  | Pole (Height h)
               \ |
                \|________ Shadow S2

Because sun rays hit the earth at approximately the same angle $\theta$ in nearby locations at the same time of day:

• The angle of elevation of the sun is equal for both objects: $\angle C = \angle F = \theta$.
• Both the pole and the tower stand perpendicular to the ground: $\angle B = \angle E = 90^\circ$.

By AA Similarity, $\triangle ABC \sim \triangle DEF$. $\frac{\text{Height of Tower } (H)}{\text{Height of Pole } (h)} = \frac{\text{Shadow of Tower } (S_1)}{\text{Shadow of Pole } (S_2)} \implies H = h \times \frac{S_1}{S_2}$

2. Cartography, Scale Models, and Blueprints

Architects design floor plans using scale ratios (e.g., $1 : 100$). Every triangular structural component in the blueprint is geometrically similar to the actual construction frame. BPT ensures that intermediate support beams placed parallel to the main wall divide the structural frame in exact proportion, guaranteeing stability.

3. Computer Graphics and Optical Zooming

When image processing software scales a digital 3D model onto a 2D monitor screen, geometric projection relies on similar triangles. The camera lens acts as a vertex point, and the object and its screen projection form similar triangles. Perspective scaling preserves aspect ratios using the similarity relation $\frac{x_{\text{screen}}}{x_{\text{world}}} = \frac{f}{z}$, where $f$ is the focal length and $z$ is the distance to the object.

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Step-by-Step Solved Textbook Examples

Example 1: Solving Algebraic Variables using BPT

Question: In $\triangle ABC$, $DE \parallel BC$. The side lengths are given as $AD = x$, $DB = x - 2$, $AE = x + 2$, and $EC = x - 1$. Find the value of $x$.

Solution:

• Step 1: Identify given conditions and state the applicable theorem. In $\triangle ABC$, we are given $DE \parallel BC$. By the Basic Proportionality Theorem (BPT): $\frac{AD}{DB} = \frac{AE}{EC}$

• Step 2: Substitute the algebraic expressions. $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$

• Step 3: Cross-multiply to clear the denominators. $x(x - 1) = (x + 2)(x - 2)$

• Step 4: Expand both sides. $x^2 - x = x^2 - 4$

• Step 5: Solve for $x$. Subtract $x^2$ from both sides: $-x = -4 \implies x = 4$

• Step 6: Verify the solution.

  • $AD = 4$
  • $DB = 4 - 2 = 2 \implies \frac{AD}{DB} = \frac{4}{2} = 2$
  • $AE = 4 + 2 = 6$
  • $EC = 4 - 1 = 3 \implies \frac{AE}{EC} = \frac{6}{3} = 2$

  Since both ratios equal $2$, $x = 4$ is valid.

Final Answer: The value of $x$ is $4$.

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Example 2: Trapezium Diagonal Ratio Proof

Question: $ABCD$ is a trapezium with $AB \parallel DC$. Diagonals $AC$ and $BD$ intersect each other at point $O$. Using a similarity criterion (or BPT), prove that: $\frac{AO}{BO} = \frac{CO}{DO}$

    A-----------B
     \         /
      \   O   /
       \     /
        C---D

Solution:

• Step 1: Identify relevant triangles formed by the intersecting diagonals. Consider $\triangle AOB$ and $\triangle COD$.

• Step 2: Identify equal angle pairs using parallel lines. Since $AB \parallel DC$ and $AC$ is a transversal: $\angle OAB = \angle OCD \quad \text{(Alternate interior angles)}$

  Since $AB \parallel DC$ and $BD$ is a transversal: $\angle OBA = \angle ODC \quad \text{(Alternate interior angles)}$

  Additionally, $\angle AOB = \angle COD$ (Vertically opposite angles).

• Step 3: Apply the Similarity Criterion. By AA Similarity Criterion, we have: $\triangle AOB \sim \triangle COD$

• Step 4: Set up the ratio of corresponding sides. Since corresponding sides of similar triangles are proportional: $\frac{AO}{CO} = \frac{BO}{DO}$

• Step 5: Rearrange terms to match the required format. Cross-multiplying or swapping the inner terms gives: $\frac{AO}{BO} = \frac{CO}{DO}$

(Hence Proved)

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Example 3: Right Triangle Perpendicularity and Altitude Relation

Question: In a right-angled triangle $ABC$, right-angled at $C$, let $p$ be the length of the perpendicular from $C$ to $AB$. If $a, b, c$ denote the lengths of the sides opposite to $\angle A, \angle B, \angle C$ respectively, prove that: $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

         C
        /|\
       / | \
      /  |p \
     /___|___\
    A    D    B

Solution:

• Step 1: Express area in two different ways. Let $CD \perp AB$, so $CD = p$. The hypotenuse is $AB = c$, base $BC = a$, and height $AC = b$.

  $\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times c \times p$ $\text{Area}(\triangle ABC) = \frac{1}{2} \times a \times b$

  Equating the two expressions for area: $\frac{1}{2} c p = \frac{1}{2} a b \implies c p = a b \implies c = \frac{a b}{p} \quad \text{--- (1)}$

• Step 2: Apply Pythagoras Theorem to $\triangle ABC$. Since $\triangle ABC$ is right-angled at $C$: $c^2 = a^2 + b^2 \quad \text{--- (2)}$

• Step 3: Substitute equation (1) into equation (2). $\left(\frac{a b}{p}\right)^2 = a^2 + b^2$ $\frac{a^2 b^2}{p^2} = a^2 + b^2$

• Step 4: Divide both sides by $a^2 b^2$. $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}$ $\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}$ $\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$

  Rearranging: $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

(Hence Proved)

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Example 4: SAS Similarity Proof

Question: In the figure below, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

           T
          / \
         /   \
        P-----\
       / \     \
      / 1 \ 2   \
     Q-----S-----R

Solution:

• Step 1: Use the given angle relation to simplify terms. In $\triangle PQR$, we are given $\angle 1 = \angle 2$ (i.e., $\angle PQR = \angle PRQ$). Since sides opposite to equal angles are equal: $PR = PQ \quad \text{--- (1)}$

• Step 2: Substitute $PR = PQ$ into the given ratio. The given ratio is: $\frac{QR}{QS} = \frac{QT}{PR}$

  Replacing $PR$ with $PQ$: $\frac{QR}{QS} = \frac{QT}{PQ}$

  Taking reciprocals/rearranging for $\triangle PQS$ and $\triangle TQR$: $\frac{PQ}{QT} = \frac{QS}{QR} \quad \text{--- (2)}$

• Step 3: Compare $\triangle PQS$ and $\triangle TQR$.

  1. From equation (2), the including sides are proportional: $\frac{PQ}{QT} = \frac{QS}{QR}$
  2. The angle included between these sides is common to both triangles: $\angle PQS = \angle TQR = \angle 1$

• Step 4: Apply SAS Criterion. By SAS Similarity Criterion: $\triangle PQS \sim \triangle TQR$

(Hence Proved)

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Common Student Mistakes to Avoid

1. Writing Incorrect Letter Orders in Similarity Statements

• The Mistake: Writing $\triangle ABC \sim \triangle DEF$ when the actual equal angles are $\angle A = \angle E$ and $\angle B = \angle D$.
• Why it loses marks: Writing $\triangle ABC \sim \triangle DEF$ automatically implies: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$ If vertices do not correspond, all derived side ratios will be mathematically incorrect.
• Correct Approach: Always write vertices in exact order of angle equality. If $A \leftrightarrow E$, $B \leftrightarrow D$, and $C \leftrightarrow F$, write $\triangle ABC \sim \triangle EDF$.

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2. Misapplying BPT Ratios with Whole Sides

• The Mistake: Setting up BPT as $\frac{AD}{AB} = \frac{AE}{EC}$.
• Why it loses marks: BPT states that a line parallel to one side divides the sides in ratio: $\frac{\text{Top Part}}{\text{Bottom Part}} = \frac{\text{Top Part}}{\text{Bottom Part}} \implies \frac{AD}{DB} = \frac{AE}{EC}$
• Correct Approach: Use consistent ratios:
  • Part-to-Part: $\frac{AD}{DB} = \frac{AE}{EC}$
  • Part-to-Whole: $\frac{AD}{AB} = \frac{AE}{AC}$
  • Bottom-to-Whole: $\frac{DB}{AB} = \frac{EC}{AC}$

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3. Assuming SAS Criterion Works with Any Equal Angle

• The Mistake: Using SAS similarity when two sides are proportional but the given equal angle is not the included angle (e.g., using $\frac{AB}{DE} = \frac{BC}{EF}$ with $\angle A = \angle D$).
• Why it loses marks: Geometric similarity under SAS requires the angle to be locked between the two proportional sides. Non-included angles can create two completely different non-similar triangles (the ambiguous case).
• Correct Approach: Verify that the equal angle is formed directly by the two proportional sides.

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4. Forgetting to State Geometric Theorems and Reasons

• The Mistake: Writing $\frac{AD}{DB} = \frac{AE}{EC}$ without providing justification.
• Why it loses marks: CBSE marking schemes assign dedicated marks for stating theorem names or conditions (e.g., "By Basic Proportionality Theorem, since $DE \parallel BC$").
• Correct Approach: Add explicit reasons in brackets beside every single deduction line.

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Practice Questions for Self-Assessment

Question 1

In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 2.4\text{ cm}$, $DB = 3.6\text{ cm}$, and $AC = 10.0\text{ cm}$, calculate the length of $AE$.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Solution:

1. Let $AE = x\text{ cm}$.
2. Since $E$ lies on $AC$, $EC = AC - AE = 10.0 - x$.
3. Since $DE \parallel BC$, by Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$
4. Substitute the values: $\frac{2.4}{3.6} = \frac{x}{10 - x}$
5. Simplify the left fraction $\frac{2.4}{3.6} = \frac{2}{3}$: $\frac{2}{3} = \frac{x}{10 - x}$
6. Cross-multiply: $2(10 - x) = 3x \implies 20 - 2x = 3x \implies 5x = 20 \implies x = 4\text{ cm}$

Final Answer: $AE = 4\text{ cm}$.

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Question 2

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground, and at the same time a nearby tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Solution:

1. Let $AB$ be the vertical pole ($6\text{ m}$) with shadow $BC = 4\text{ m}$.
2. Let $PQ$ be the vertical tower ($h\text{ m}$) with shadow $QR = 28\text{ m}$.
3. In $\triangle ABC$ and $\triangle PQR$:
   • $\angle B = \angle Q = 90^\circ$ (Both are vertical to the ground)
   • $\angle C = \angle R$ (Sun's elevation angle is identical at the same time)
4. By AA Similarity Criterion, $\triangle ABC \sim \triangle PQR$.
5. Ratio of corresponding sides: $\frac{AB}{PQ} = \frac{BC}{QR} \implies \frac{6}{h} = \frac{4}{28}$
6. Simplify $\frac{4}{28} = \frac{1}{7}$: $\frac{6}{h} = \frac{1}{7} \implies h = 6 \times 7 = 42\text{ m}$

Final Answer: The height of the tower is $42\text{ meters}$.

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Question 3

In $\triangle ABC$, $D$ is a point on side $BC$ such that $\angle ADC = \angle BAC$. Prove that $CA^2 = CB \cdot CD$.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Solution:

1. Consider $\triangle ABC$ and $\triangle DAC$.
2. Identify equal angles:
   • $\angle BAC = \angle ADC$ (Given)
   • $\angle C = \angle C$ (Common angle to both triangles)
3. By AA Similarity Criterion: $\triangle ABC \sim \triangle DAC$
4. Write the ratio of corresponding sides: $\frac{AC}{DC} = \frac{BC}{AC}$
5. Cross-multiply: $AC \cdot AC = BC \cdot DC \implies CA^2 = CB \cdot CD$

(Hence Proved)

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Question 4

In $\triangle ABC$, $AD \perp BC$ and $BD : AD = AD : CD$. Prove that $\triangle ABC$ is a right-angled triangle at $A$.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Solution:

1. We are given $\frac{BD}{AD} = \frac{AD}{CD}$, which can be rewritten as: $\frac{BD}{AD} = \frac{AD}{CD} \quad \text{--- (1)}$
2. In $\triangle BDA$ and $\triangle ADC$:
   • $\angle BDA = \angle ADC = 90^\circ$ (Since $AD \perp BC$)
   • From equation (1), the sides including these right angles are proportional: $\frac{BD}{AD} = \frac{AD}{CD}$
3. By SAS Similarity Criterion: $\triangle BDA \sim \triangle ADC$
4. Since corresponding angles of similar triangles are equal:
   • Let $\angle BAD = \angle ACD = x$
   • Let $\angle ABD = \angle CAD = y$
5. In right $\triangle BDA$, $\angle ABD + \angle BAD = 90^\circ \implies x + y = 90^\circ$.
6. Now evaluate angle $A$: $\angle BAC = \angle BAD + \angle CAD = x + y = 90^\circ$

Final Answer: $\triangle ABC$ is right-angled at $A$.

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Exam Revision & Frequently Asked Questions (FAQs)

Q1: Is writing AA similarity sufficient, or must I explicitly prove AAA similarity in board exams?

Answer: Writing AA Similarity is completely valid, fully accepted by CBSE marking schemes, and preferred for brevity. Since the sum of interior angles in any triangle is always $180^\circ$, two pairs of equal angles automatically guarantee that the third pair is equal.

Q2: How do I know whether to use BPT or Similarity Criteria when solving a geometry problem?

Answer: Use this simple guideline:

• Use BPT when a single triangle contains a line drawn parallel to one of its sides intersecting the other two sides.
• Use Similarity Criteria ($AA, SSS, SAS$) when comparing two distinct triangles or when proving relations involving all three sides of two triangles (especially lines intersecting at central vertices or diagonals crossing inside quadrilaterals).

Q3: Can BPT be directly applied inside a trapezium?

Answer: No, BPT applies strictly to triangles. To use BPT in a trapezium $ABCD$ ($AB \parallel DC$), you must first draw a diagonal (e.g., $AC$) to divide the trapezium into two triangles ($\triangle ABC$ and $\triangle ADC$), and then apply BPT to each triangle individually. Alternatively, you can directly use the AA Similarity Criterion on the vertically opposite triangles formed by both diagonals intersecting at $O$.

Q4: If two triangles are similar, is the ratio of their perimeters equal to the ratio of their corresponding sides?

Answer: Yes. If $\triangle ABC \sim \triangle DEF$ with scale factor $k$: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k$ $\implies AB = k \cdot DE, \quad BC = k \cdot EF, \quad AC = k \cdot DF$ Adding the sides together to find the perimeter ratio: $\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle DEF)} = \frac{AB + BC + AC}{DE + EF + DF} = \frac{k(DE + EF + DF)}{DE + EF + DF} = k$ Thus, the ratio of perimeters of two similar triangles always equals the ratio of any pair of corresponding sides.