All congruent figures are similar, but all similar figures are not congruent.
All congruent figures are similar, but all similar figures are not congruent.
Chapter 6 · CBSE Class X
📘 Definition
Definition
Two geometric figures are said to be similar if they have the same shape but not necessarily the same size.
This means:
Two geometric figures are said to be similar if they have the same shape but not necessarily the same size.
This means:
- All corresponding angles are equal.
- All corresponding sides are proportional.
🎨 SVG Diagram
Visual Understanding
🔢 Formula
Important Formula
- \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \)
- \( \dfrac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left( \dfrac{AB}{DE} \right)^2 \)
📌 Note
Core Properties
- Ratio of corresponding sides is constant (called scale factor).
- Corresponding angles are equal.
- Ratio of perimeters = ratio of sides.
- Ratio of areas = square of ratio of sides.
📄 Examples
Solved Example
Example 1
If two triangles are similar and the ratio of their corresponding sides is 3:5, find the ratio of their areas.
Using area property:
Using area property:
- \( \text{Ratio of areas} = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25} \)
Final Answer: 9 : 25
🔬 Proof
Derivation Insight
- When two similar triangles are scaled by a factor k, their corresponding sides become k times. Since area depends on square of dimensions:
- \( \text{Area} \propto (\text{side})^2 \Rightarrow \text{Area ratio} = k^2 \)
📌 Note
Exam Tip
- Always write corresponding sides in correct order
- Use similarity criteria (AA, SAS, SSS) before applying ratios.
- Area questions are frequently asked (2–3 marks).
- Diagram-based reasoning gives step marks.
⚠️ Warning
Common Mistakes
- Mixing up corresponding sides.
- Using linear ratio instead of squared ratio for area.
- Assuming similarity without proving criteria.
- Incorrect labeling of triangles.
📋 Case Study
CBSE Case Study (HOTS)
A model of a tower is made such that its height is 1.5 m, while the actual tower is 30 m high.
If the base area of the model is 200 cm², find the base area of the actual tower.
- Scale factor = \( \dfrac{30}{1.5} = 20 \)
- Area scale factor = \( 20^2 = 400 \)
- Actual area = \( 200 \times 400 = 80,000 \, \text{cm}^2 \)
Answer: 80,000 cm²
📘 Definition
Similarity of Triangles
Two triangles are said to be similar if:
- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio (proportional).
\[ Symbolically: \triangle ABC \sim \triangle DEF \]
📄 Visual
Visual Understanding
Visual Representation
🔢 Formula
Key Results & Formulas
- \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \)
- Ratio of areas = square of ratio of sides
- \( \dfrac{\text{Area}_1}{\text{Area}_2} = \left( \dfrac{Side_1}{Side_2} \right)^2 \)
📄 Examples
Solved Example
Example 1
In triangles ABC and DEF, ∠A = ∠D and ∠B = ∠E. Prove similarity.
- Since two angles are equal:
- \( \triangle ABC \sim \triangle DEF \) (AA criterion)
Conclusion: Triangles are similar.
📌 Note
Exam Tip
- Always state the criterion (AA/SAS/SSS) clearly.
- Maintain correct order while writing similarity: ABC ~ DEF.
- Diagram labeling must match written correspondence.
- Proof-based questions carry 2–4 marks frequently.
⚠️ Warning
Common Mistakes
- Wrong correspondence (e.g., mixing A with E instead of D).
- Using SSS without checking proportionality properly.
- Skipping justification of similarity criterion.
- Confusing similarity with congruence.
📋 Case Study
CBSE Case Study (HOTS)
A shadow of a tree and a pole are observed at the same time. The tree casts a 10 m shadow, while a 2 m pole casts a 1 m shadow. Find the height of the tree.
- By similarity:
- \( \dfrac{\text{Height of tree}}{2} = \dfrac{10}{1} \)
- Height = \( 20 \, m \)
Answer: 20 m
🌟 Importance
Importance for Board Exams
- Core foundation of entire chapter (appears in almost every question).
- Used in proofs, constructions, and applications.
- Direct link to Basic Proportionality Theorem.
- High scoring with predictable patterns.
🖼️ Figure
Thales of Miletus
📘 Definition
Basic Proportionality Theorem (BPT) / Thales' Theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
🎨 SVG Diagram
Diagram — BPT Triangle
Basic Proportionality Theorem (BPT)
🔢 Formula
Mathematical Form
- In △ABC, if DE ∥ BC, then:
- \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \)
💡 Concept
Concept Simplified
A line parallel to one side ensures that the triangle is divided into two smaller triangles that are similar to the original triangle. Due to similarity, corresponding sides maintain equal ratios.
🔬 Proof
Derivation Insight
In △ABC, DE ∥ BC:
- \(\angle ADE = \angle ABC (alternate interior angles)\)
- \(\angle AED = \angle ACB\)
- \(\Rightarrow \triangle ADE \sim \triangle ABC (AA similarity)\)
- \(\Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC} \)
- \(\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \)
✏️ Example
Solved Example
Example 2
In a triangle, a line parallel to one side divides the other two sides in the ratio 2:3.
If one segment is 4 cm, find the other segment.
- Let ratio = 2 : 3
- If smaller part = 4 cm ⇒ scale factor = 2
- Larger part = \( 3 \times 2 = 6 \) cm
- Answer: 6 cm
📌 Note
Exam Tip
- Always mention DE ∥ BC before applying BPT.
- Use similarity before writing proportionality.
- Steps must include reason (angles, similarity, ratio).
- Frequently asked in 2–3 mark proof questions.
⚠️ Warning
Common Mistakes
- Applying BPT without checking parallel condition.
- Writing incorrect ratios (wrong order).
- Skipping similarity step in proofs.
- Confusing with converse theorem.
📋 Case Study
CBSE Case Study (HOTS)
A road cuts two sides of a triangular field parallel to the third side. It divides one side in ratio 3:5.
Find the ratio in which the other side is divided.
- By BPT:
- Same ratio applies ⇒ 3 : 5
- Answer: 3 : 5
🌟 Importance
Importance for Board Exams
- Foundation for triangle similarity and constructions.
- Used in coordinate geometry and real-life scaling problems.
- Very high probability in CBSE exams.
- Direct concept-based scoring (easy marks if understood).
📘 Definition
Theorem 1: Basic Proportionality Theorem (Proof)
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,
then the other two sides are divided in the same ratio.
🔬 Proof
Derivation Insight
Given
In △ABC, DE ∥ BC, where D lies on AB and E lies on AC.
- To Prove
- \[\dfrac{AD}{DB} = \dfrac{AE}{EC}\]
Construction
- Join BE and CD.
- Draw DM ⟂ AC and EN ⟂ AB.
- Proof
- Area of triangle ADE:
- \[ar(ADE) = \dfrac{1}{2} \times AD \times EN\]
- Area of triangle BDE:
- \[ar(BDE) = \dfrac{1}{2} \times BD \times EN\]
- \[\dfrac{ar(ADE)}{ar(BDE)} = \dfrac{AD}{DB} \tag{1}\]
- Also, using altitude DM:
- \[ar(ADE) = \dfrac{1}{2} \times AE \times DM\]
- \[ar(DEC) = \dfrac{1}{2} \times EC \times DM\]
- Dividing
- \[\dfrac{ar(ADE)}{ar(DEC)} = \dfrac{AE}{EC} \tag{2}\]
- Triangles BDE and DEC lie on the same base DE and between the same parallels BC and DE.
- \[ar(BDE) = ar(DEC) \tag{3}\]
- From (1), (2), and (3):
- \[\boxed{\bbox[indigo,5pt]{\dfrac{AD}{DB} = \dfrac{AE}{EC}}}\]
🌟 Importance
Key Insight
The proof relies on the principle that triangles on the same base and between the same parallels
have equal areas. This bridges area geometry with proportional reasoning.
📌 Note
Exam Tip
- Always mention same base & same parallels ⇒ equal areas.
- Tag equations (1), (2), (3) for clarity and marks.
- Diagram labeling must match steps exactly.
- This proof is frequently asked for 3–4 marks.
⚠️ Warning
Common Mistakes
- Writing \( \dfrac{AE}{AC} \) instead of \( \dfrac{AE}{EC} \).
- Skipping construction (DM ⟂ AC, EN ⟂ AB).
- Not justifying equal areas of triangles.
- Improper step linking between equations.
📘 Definition
Theorem 2: Converse of Basic Proportionality Theorem
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
🖼️ Figure
Converse of BPT Diagram
🔬 Proof
Derivation
- In \(\triangle \)ABC, points D and E lie on AB and AC respectively such that: \[ \dfrac{AD}{DB} = \dfrac{AE}{EC} \tag{1} \]
- To Prove
- \[DE \parallel BC\]
Construction
- Through point D, draw a line parallel to BC intersecting AC at E.
- Proof
- Since \( DE' \parallel BC \), by Basic Proportionality Theorem:
- \[\dfrac{AD}{DB} = \dfrac{AE'}{E'C} \tag{2}\]
- From (1) and (2):
- \[\dfrac{AE}{EC} = \dfrac{AE'}{E'C} \tag{3}\]
- Adding 1 to both sides:
- \[\dfrac{AE + EC}{EC} = \dfrac{AE' + E'C}{E'C}\]
- \[\Rightarrow \dfrac{AC}{EC} = \dfrac{AC}{E'C}\]
- This is possible only when:
- \[EC = E'C\]
- Hence, point E coincides with E'. Therefore, the line DE is the same as DE'.
🌟 Importance
Key Insight
The proof uses the method of contradiction and uniqueness of division point.
If two ratios are equal, the division point on a line is unique, forcing parallelism.
📌 Note
Exam Tip
- Always construct a parallel line (DE').
- Apply BPT first, then compare ratios.
- Clearly state “E coincides with E'”.
- This theorem is frequently asked with reasoning steps (3–4 marks).
⚠️ Warning
Common Mistakes
- Skipping construction of E'.
- Not proving uniqueness of point.
- Jumping directly to conclusion without ratio comparison.
- Incorrect algebra while adding ratios.
✏️ Example
Quick Example
Example 3
- If a line divides sides of a triangle in ratio 4:5 on both sides, then the line is parallel to the third side.
- Conclusion: Converse of BPT applies ⇒ line is parallel.
✏️ Example
Solved Example
Example 4
If a line intersects sides \(AB\) and \(AC\) of a triangle \(ABC\) at points \(D\) and \(E\) respectively
and is parallel to \(BC\), prove that:
\[ \dfrac{AD}{AB} = \dfrac{AE}{AC} \]
🖼️ Figure
Fig. 6.13
🔬 Proof
To Prove \[\dfrac{AD}{AB} = \dfrac{AE}{AC}\]
Proof
Given
In △ABC, D lies on AB and E lies on AC such that:
\[DE \parallel BC\]
- Since \( DE \parallel BC \), by Basic Proportionality Theorem:
- \[\dfrac{AD}{DB} = \dfrac{AE}{EC}\]
- Taking reciprocal:
- \[\dfrac{DB}{AD} = \dfrac{EC}{AE} \tag{1}\]
- Adding 1 to both sides of (1):
- \[\dfrac{DB}{AD} + 1 = \dfrac{EC}{AE} + 1\]
- \[\Rightarrow \dfrac{DB + AD}{AD} = \dfrac{EC + AE}{AE}\]
- Using segment addition:
- \[DB + AD = AB \quad \text{and} \quad EC + AE = AC\]
- \[\Rightarrow \dfrac{AB}{AD} = \dfrac{AC}{AE}\]
- Taking reciprocal:
- \[\boxed{\bbox[indigo,5pt]{\dfrac{AD}{AB} = \dfrac{AE}{AC}}}\]
🌟 Importance
Key Insight
This result is a derived proportional form of BPT. It shows that ratios can be
manipulated algebraically to match required proof formats.
📌 Note
Exam Tip
- Always start with BPT before transforming ratios.
- Clearly show “taking reciprocal” and “adding 1” steps.
- Use segment addition explicitly (AB = AD + DB).
- Final step must match exactly what is asked.
⚠️ Warning
Common Mistakes
- Skipping reciprocal step.
- Incorrect addition of segments.
- Jumping directly to final result.
- Writing wrong final ratio order.
✏️ Example
Solved Example : BPT in a Trapezium
Example 5
ABCD is a trapezium with \( AB \parallel DC \). Points E and F lie on AD and BC respectively such that
\( EF \parallel AB \). Show that:
\[ \dfrac{AE}{ED} = \dfrac{BF}{FC} \]
📖 Theory
Theory + Roadmap (How to Think)
Core Strategy: Diagonal + Parallel Lines ⇒ Apply BPT twice ⇒ Equate ratios
- When a trapezium is involved → draw diagonal AC
- Parallel lines suggest → use Basic Proportionality Theorem (BPT)
- Break figure into two triangles:
- \( \triangle ADC \)
- \( \triangle ABC \)
- Find a common ratio (AG/GC) using both triangles.
- Equate results → get required proof.
🖼️ Figure
Fig. 6.14
🔬 Proof
To Prove
\[ \dfrac{AE}{ED} = \dfrac{BF}{FC} \]
🔬 Proof
Given
\[ AB \parallel DC,\quad EF \parallel AB \]
🔬 Proof
Construction
Join diagonal AC. Let EF intersect AC at point G.
🔬 Proof
Proof
- In \( \triangle ADC \):
- Since \( EG \parallel DC \) (as \( EF \parallel DC \)), by BPT:
- \[\dfrac{AE}{ED} = \dfrac{AG}{GC} \tag{1}\]
- In \( \triangle ABC \):
- Since \( GF \parallel AB \) (as \( EF \parallel AB \)), by BPT:
- \[\dfrac{BF}{FC} = \dfrac{AG}{GC} \tag{2}\]
- From (1) and (2):
- \[\boxed{\bbox[indigo,5pt]{\dfrac{AE}{ED} = \dfrac{BF}{FC}}}\]
🌟 Importance
Key Insight
The diagonal \( AC \) acts as a bridge connecting two triangles where BPT can be applied.
This is a standard CBSE technique for trapezium problems.
📌 Note
Exam Tip
- Always draw diagonal when dealing with trapezium.
- Identify two triangles where BPT can be applied.
- Clearly mention parallel lines before applying BPT.
- Equate common ratios (AG/GC) to conclude.
⚠️ Warning
Common Mistakes
- Not defining point G (intersection on AC).
- Writing incorrect final ratio (AE/AD instead of AE/ED).
- Confusing which lines are parallel.
- Skipping triangle identification step.
📘 Definition
Criteria for Similarity of Triangles (AA, SAS, SSS)
Example 6
Questions (Exam Pattern)
- In two triangles, two angles are equal. Prove that the triangles are similar.
- Two sides of two triangles are proportional and the included angle is equal. Are the triangles similar?
- The sides of two triangles are in the ratio 2:3, 4:6, and 6:9. Check similarity.
- Identify which similarity criterion (AA, SAS, SSS) applies in each case.
📖 Theory
Theory + Roadmap (How to Decide Criterion)
- Angles given? → Use AA
- Two sides + included angle? → Use SAS
- All sides proportional? → Use SSS
Decision Shortcut: Count known elements → Match pattern → Apply correct criterion
🎨 SVG Diagram
Visual Understanding
💡 Concept
Similarity Criteria Explained
-
AA (Angle–Angle):
If two angles of one triangle are equal to two angles of another triangle, then triangles are similar.
Reason: Third angle automatically becomes equal. -
SAS (Side–Angle–Side):
If two sides are proportional and the included angle is equal, then triangles are similar. -
SSS (Side–Side–Side):
If all three corresponding sides are proportional, then triangles are similar.
✏️ Example
Example 5
Solved
Solutions
Q1: Two angles equal ⇒ AA similarity ⇒ Triangles are similar.
Q2: Two sides proportional + included angle equal ⇒ SAS similarity.
Q3: Ratios:
\[ \dfrac{2}{3} = \dfrac{4}{6} = \dfrac{6}{9} \]
⇒ All sides proportional ⇒ SSS similarity.
Q4: Identify based on given data using roadmap above.
⚡ Exam Tip
Exam Tips
- Always state the criterion name (AA, SAS, SSS).
- Maintain correct order of vertices.
- Justification step is compulsory for full marks.
- AA is most frequently used in proofs.
⚠️ Warning
Common Mistakes
- Using SAS without included angle.
- Incorrect ratio comparison in SSS.
- Not checking correspondence order.
- Forgetting to mention criterion name.
📘 Definition
Question
AA Similarity Criterion
In two triangles, if corresponding angles are equal, prove that their corresponding sides are proportional
and hence the triangles are similar.
📖 Theory
Theory + Roadmap
- Given angles equal → think AA similarity
- Construct a triangle equal in size using given sides.
- Use congruency (SAS) to match triangles.
- Then apply BPT to prove proportionality.
Strategy: Equal angles → Construct equal sides → Use congruency → Apply BPT → Get ratios
🖼️ Figure
AA Similarity Theorem Diagram
🔬 Proof
Given
In triangles ABC and DEF:
\[ \angle A = \angle D,\quad \angle B = \angle E,\quad \angle C = \angle F \]
🔬 Proof
To Prove
\[ \begin{aligned}\dfrac{AB}{DE} = \dfrac{BC}{EF} &= \dfrac{AC}{DF}\quad\text{ and hence}\\\\ \triangle ABC &\sim \triangle DEF\end{aligned} \]
🔬 Proof
Construction
- On DE, mark point P such that \( DP = AB \).
- On DF, mark point Q such that \( DQ = AC \).
- Join PQ.
🔬 Proof
Proof
⇒ \( \triangle ABC \cong \triangle DPQ \) (SAS)
⇒ \( \angle B = \angle P,\quad \angle C = \angle Q \) (CPCT)
Step 2: Parallel Lines
Given \( \angle B = \angle E \), and from above \( \angle B = \angle P \),
⇒ \( \angle P = \angle E \)
Similarly, \( \angle Q = \angle F \)
⇒ \( PQ \parallel EF \)
Step 3: Apply BPT
In triangle DEF, since \( PQ \parallel EF \):
\[ \dfrac{DP}{PE} = \dfrac{DQ}{QF} \]Substituting \( DP = AB \) and \( DQ = AC \):
\[ \dfrac{AB}{DE} = \dfrac{AC}{DF} \]Similarly,
\[ \dfrac{AB}{DE} = \dfrac{BC}{EF} \]Hence,
\[ \boxed{\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}} \]⇒ \( \triangle ABC \sim \triangle DEF \)
Congruency
In \(\triangle\) ABC and \(\triangle\) DPQ:
In \(\triangle\) ABC and \(\triangle\) DPQ:
- \(DP = AB \) (construction)
- \(DQ = AC \) (construction)
- \(\angle PDQ = \angle A \)
📄 Important
Key Insight
This proof shows that angle equality alone guarantees shape similarity.
Side proportionality is a consequence, not a requirement.
⚡ Exam Tip
Exam Tips
- AA is the most used similarity criterion in CBSE.
- Always justify parallel lines using equal angles.
- Clearly mention SAS and BPT steps.
- Maintain correct triangle correspondence order.
⚠️ Warning
Common Mistakes
- Skipping construction steps.
- Not proving parallelism before BPT.
- Incorrect correspondence of vertices.
- Jumping directly to similarity without ratios.
📄 Result
If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar (AA Similarity Criterion).
📄 Label
Question
If in two triangles, the corresponding sides are proportional, prove that their corresponding angles are equal
and hence the triangles are similar.
📖 Theory
Theory + Roadmap
- Given all three sides proportional → think SSS similarity
- Create a triangle with equal sides using construction.
- Use congruency (SSS) to match shapes.
- Then compare angles → prove similarity.
Strategy: Proportional sides → Construct equal triangle → Use congruency → Conclude similarity
🖼️ Figure
SSS Similarity Theorem Diagram
🔬 Proof
Given
In triangles ABC and DEF:
\[ \begin{aligned}\angle A = \angle D,\ \angle B &= \angle E,\ \angle C = \angle F\\\\ \Rightarrow \triangle ABC &\sim \triangle DEF\end{aligned} \]
🔬 Proof
To Prove
\[ \begin{aligned}\angle A = \angle D,\ \angle B &= \angle E,\ \angle C = \angle F\\\\ \Rightarrow \triangle ABC &\sim \triangle DEF\end{aligned} \]
🔬 Proof
- On DE, mark point P such that \( DP = AB \).
- On DF, mark point Q such that \( DQ = AC \).
- Join PQ.
🔬 Proof
Proof
Step 1: Use Given Ratios
Given:
\[ \dfrac{AB}{DE} = \dfrac{AC}{DF} \]Substitute \( DP = AB \) and \( DQ = AC \):
\[ \dfrac{DP}{DE} = \dfrac{DQ}{DF} \]⇒ \( \dfrac{DP}{DE} = \dfrac{DQ}{DF} \)
⇒ By Converse of BPT, \( PQ \parallel EF \)
Step 2: Angle Equality
Since \( PQ \parallel EF \):
- \( \angle DPQ = \angle E \)
- \( \angle DQP = \angle F \)
Step 3: Compare Triangles
In triangles ABC and DPQ:
- \( DP = AB \)
- \( DQ = AC \)
- \( \angle PDQ = \angle A \)
⇒ \( \triangle ABC \cong \triangle DPQ \) (SAS)
⇒ \( \angle A = \angle D,\ \angle B = \angle E,\ \angle C = \angle F \)
⇒ \( \triangle ABC \sim \triangle DEF \)
📌 Note
Key Insight
📄 Tips
Exam Tips
- Always start with given ratios.
- Use Converse of BPT carefully.
- Clearly justify parallel lines.
- Write final similarity statement explicitly.
⚠️ Warning
Common Mistakes
- Using BPT instead of Converse of BPT.
- Incorrect substitution of constructed lengths.
- Skipping parallel proof step.
- Confusing congruency and similarity.
📄 Result
Final Result
If the corresponding sides of two triangles are proportional, then the triangles are similar
(SSS Similarity Criterion).
📄 Label
Question
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles
are proportional, prove that the two triangles are similar.
📖 Theory
Theory + Roadmap
- Given: one angle equal + two sides proportional → think SAS similarity.
- Construct equal sides to form a comparable triangle.
- Use congruency (SAS) to match triangles.
- Then extend result to full triangle → prove similarity.
Strategy: Proportional sides + included angle → Construct → Congruency → Extend → Similarity
🖼️ Figure
SAS Similarity Theorem Diagram
🔬 Proof
Given
In triangles ABC and DEF:
\[ \dfrac{AB}{DE} = \dfrac{AC}{DF}, \quad \angle A = \angle D \]
🔬 Proof
To Prove
\[ \triangle ABC \sim \triangle DEF \]
🔬 Proof
- On DE, mark point P such that \( DP = AB \).
- On DF, mark point Q such that \( DQ = AC \).
- Join PQ.
🔬 Proof
Proof
Step 1: Use Given Ratio
\[ \dfrac{AB}{DE} = \dfrac{AC}{DF} \]Substitute \( DP = AB \), \( DQ = AC \):
\[ \dfrac{DP}{DE} = \dfrac{DQ}{DF} \]⇒ By Converse of BPT, \( PQ \parallel EF \)
Step 2: Angle Equality
Since \( PQ \parallel EF \):
- \( \angle DPQ = \angle E \)
- \( \angle DQP = \angle F \)
Step 3: Congruency
In triangles ABC and DPQ:
- \( DP = AB \)
- \( DQ = AC \)
- \( \angle PDQ = \angle A \)
⇒ \( \triangle ABC \cong \triangle DPQ \) (SAS)
⇒ \( \angle B = \angle P,\ \angle C = \angle Q \)
Step 4: Final Comparison
From above:
- \( \angle B = \angle E \)
- \( \angle C = \angle F \)
⇒ All corresponding angles equal, therfore
\[ \boxed{\bbox[indigo,5pt]{\triangle ABC \sim \triangle DEF}} \]
📌 Note
Key Insight
📄 Tips
Exam Tips
- Always verify the angle is included angle
- Use Converse of BPT carefully.
- Clearly mention parallel lines before angle comparison.
- SAS is often used in coordinate geometry applications.
⚠️ Warning
Common Mistakes
- Using non-included angle.
- Skipping construction step.
- Wrong use of BPT instead of Converse.
- Incorrect angle correspondence.
📄 Result
Final Result
If one angle is equal and the including sides are proportional, then triangles are similar
(SAS Similarity Criterion).
📄 Label
Question
In Fig. 6.29, if \( PQ \parallel RS \), prove that:
\[ \triangle POQ \sim \triangle SOR \]
📖 Theory
Theory + Roadmap
- Parallel lines → think alternate interior angles
- Intersecting lines → give vertically opposite angles
- Two equal angles are enough → use AA similarity
Strategy: Parallel lines → angle equality → apply AA similarity
🖼️ Figure
Parallel Lines Triangle Similarity Diagram
🔬 Proof
Solution
Given
\[ PQ \parallel RS \]
🔬 Proof
To Prove
\[ \triangle POQ \sim \triangle SOR \]
🔬 Proof
Proof
In triangles \( \triangle POQ \) and \( \triangle SOR \):
- \( \angle OPQ = \angle OSR \) (Alternate interior angles, since \( PQ \parallel RS \))
- \( \angle OQP = \angle ORS \) (Alternate interior angles)
- \( \angle POQ = \angle SOR \) (Vertically opposite angles)
Since corresponding angles are equal:
\[ \triangle POQ \sim \triangle SOR \quad \text{(AA similarity criterion)} \]
📌 Note
Key Insight
📄 Tips
Exam Tips
- Always name angles properly (e.g., \( \angle OPQ \), not just \( \angle P \))
- Two angles are sufficient → no need to prove all three.
- Always mention AA similarity explicitly.
⚠️ Warning
Common Mistakes
- Writing incomplete angles (like ∠P instead of ∠OPQ).
- Not mentioning reason (alternate / vertically opposite).
- Skipping similarity criterion name.
📄 Label
Quick Check (Exam Revision)
- When are two figures similar?
- What is the difference between similarity and congruence?
- Which conditions define similarity of polygons?
- When can BPT and its converse be applied?
- How do AA, SAS, and SSS criteria differ?
📖 Theory
Concept Roadmap
- Shape same → Similarity
- Shape + size same → Congruence
- Parallel lines → BPT
- Angles equal → AA
- Sides proportional → SSS
- Sides + included angle → SAS
📄 Label
Core Theoretical Points
- Two figures having the same shape but not necessarily the same size are called similar figures.
- All congruent figures are similar, but similar figures are not necessarily congruent.
- Two polygons are similar if:
- Corresponding angles are equal
- Corresponding sides are proportional
- Basic Proportionality Theorem (BPT):
A line parallel to one side of a triangle divides the other two sides in the same ratio. - Converse of BPT:
If a line divides two sides in the same ratio, then it is parallel to the third side. - SSS (Side–Side–Side) Similarity:
All corresponding sides proportional ⇒ triangles are similar. - SAS (Side–Angle–Side) Similarity:
Two sides proportional + included angle equal ⇒ triangles are similar.
🔢 Formula
Important Formulas
- \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \)
- \( \dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{Side}_1}{\text{Side}_2}\right)^2 \)
📄 Tips
Exam Tips
- Always mention the similarity criterion name
- Check correspondence order carefully
- Use diagrams for better clarity and step marks.
- BPT and similarity are frequently linked in questions.
⚠️ Warning
Common Mistakes
- Confusing similarity with congruence.
- Using SAS without included angle.
- Applying BPT without parallel condition.
- Incorrect ratio formation.
⚡ Exam Tip
Final Tip
💡 Concept
NCERT Mathematics · Class X · Chapter 6
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