Class 10 Triangles Notes

Exam Weightage & Blueprint

Total: 7-9 Marks

This chapter falls under Unit IV: Geometry (15 marks total). As per the latest syllabus, focus is on: proof and application of the Basic Proportionality Theorem (BPT) and criteria for similarity of triangles (AAA, SSS, SAS).

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Basic Proportionality Theorem (BPT) Applications
Short Answer	2 or 3	Medium	Similarity Criteria (AA, SAS, SSS)
Long Answer	5	High	Proof of BPT or Similarity Theorems

⏰ Last 24-Hour Checklist

Congruent vs Similar: Similar = same shape, diff size.
BPT (Thales Theorem): $\frac{AD}{DB} = \frac{AE}{EC}$ if $DE || BC$.
Similarity Criteria: AAA, AA, SSS, SAS.

Area Ratio Theorem: Ratio of areas = Ratio of squares of corresp. sides (Deleted in some boards, check syllabus).
Shadow Problems: Use similarity of triangles formed by sun rays.

📐 Theorems & Proofs

Basic Proportionality Theorem (BPT / Thales Theorem – 5 Marks): 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.

Full Step-by-Step Board Proof of BPT (Theorem 6.1)

Given: A triangle $ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively ($DE \parallel BC$).

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

Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.

Figure: Basic Proportionality Theorem Construction & Altitudes

Step 1: Area of $\triangle ADE = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$

Step 2: Area of $\triangle BDE = \frac{1}{2} \times DB \times EN \quad (\text{height is altitude } EN \text{ on extended side } DB)$

Taking ratio: $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)}$

Step 3: Similarly, taking base $AE$ and altitude $DM$ on $AC$:

$\text{ar}(\triangle ADE) = \frac{1}{2} \times AE \times DM$, $\text{ar}(\triangle DEC) = \frac{1}{2} \times EC \times DM$

$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)}$

Step 4: $\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $DE$ and $BC$.

$\therefore \text{ar}(\triangle BDE) = \text{ar}(\triangle DEC) \quad \text{--- (3)}$

From (1), (2), and (3):

$$ \frac{AD}{DB} = \frac{AE}{EC} \quad \text{[Hence Proved]} $$

🎯 Important Corollaries of BPT (Used Heavily in MCQs & 2-Mark Questions):

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

Converse of BPT (Theorem 6.2): If a line divides any two sides of a triangle in the same ratio ($\frac{AD}{DB} = \frac{AE}{EC}$), then the line is parallel to the third side ($DE \parallel BC$).

Internal Angle Bisector Theorem (Competitive/Optional Edge): The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle: $\frac{BD}{DC} = \frac{AB}{AC}$.

📐 Similarity Criteria

AAA / AA

If corresponding angles are equal, triangles are similar.

SSS

If corresponding sides are proportional, triangles are similar.

SAS

One angle equal and sides including it are proportional.

Note: AA is sufficient for similarity. You don't need to prove all three angles.

ΔABC ∼ ΔDEF: Equiangular & Corresponding Sides are in Proportion

Solved Examples (Board Marking Scheme)

Q1. In $\triangle ABC$, $DE || BC$. If $AD=1.5$ cm, $DB=3$ cm, $AE=1$ cm, find $EC$. (2 Marks)

Step 1: State Theorem 0.5 Mark

Since $DE || BC$, by Basic Proportionality Theorem:

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

Step 2: Substitute Values 1 Mark

$\frac{1.5}{3} = \frac{1}{EC}$

$\frac{1}{2} = \frac{1}{EC}$

Step 3: Solve 0.5 Mark

$EC = 2$ cm.

Q2. Diagonals of a trapezium ABCD with $AB || DC$ intersect at O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$. (3 Marks)

Step 1: Identify Triangles 1 Mark

Consider $\triangle AOB$ and $\triangle COD$.

$\angle AOB = \angle COD$ (Vertically opposite angles)

Step 2: Alternate Angles 1 Mark

Since $AB || DC$, alternate interior angles are equal:

$\angle OAB = \angle OCD$

Step 3: Apply Similarity 1 Mark

By AA criterion, $\triangle AOB \sim \triangle COD$.

Therefore, corresponding sides are proportional:

$\frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$ (Rearranging terms)

Previous Year Questions (PYQs)

2023 (1 Mark): If $\triangle ABC \sim \triangle DEF$ and $AB=3, DE=4.5$, find ratio of areas.
Ans: Ratio of areas = Square of ratio of sides = $(3/4.5)^2 = (2/3)^2 = 4:9$.

2020 (3 Marks): A vertical pole of length 6m casts a shadow 4m long. At the same time, a tower casts a shadow 28m long. Find height of tower.
Ans: $\triangle ABC \sim \triangle PQR$ (Sun's elevation is same). $\frac{6}{h} = \frac{4}{28} \Rightarrow h = 42$ m.

2019 (4 Marks): Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Ans: Standard theorem proof. Draw altitudes, use area formula and similarity.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Order of Vertices: Writing $\triangle ABC \sim \triangle DEF$ when actually $\triangle ABC \sim \triangle EDF$. Order matters!

BPT vs Similarity: Confusing $\frac{AD}{DB} = \frac{AE}{EC}$ (BPT) with $\frac{AD}{AB} = \frac{DE}{BC}$ (Similarity).

Drawing Figures: Not drawing a figure for geometry questions. It is mandatory for marks.

💡 Scoring Tips

Correspondence: Always check which angle corresponds to which before writing similarity.

Given/To Prove: Always write "Given", "To Prove", and "Construction" headings in theorem proofs.

Shadow Problems: Remember "at the same time" means the angle of elevation of the sun is equal for both objects.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): All equilateral triangles are ___________ (congruent / similar).

Q2 (2M): In $\triangle PQR$, $S$ and $T$ are points on $PQ$ and $PR$ respectively such that $ST \parallel QR$. If $PS = 3$ cm, $SQ = 3$ cm, and $PT = 4$ cm, find $PR$.

Q3 (3M): A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Q4 (4M): State and prove the Basic Proportionality Theorem (Thales Theorem).

📝 More Solved Board Questions

Q3. In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$, $\angle B = \angle E$, $AB = 3$ cm, $BC = 5$ cm, and $DE = 6$ cm. Find $EF$. 2 Marks

Sol. Since two angles are equal, $\triangle ABC \sim \triangle DEF$ by AA Similarity Criterion.

Therefore, $\frac{AB}{DE} = \frac{BC}{EF}$

$\frac{3}{6} = \frac{5}{EF} \Rightarrow \frac{1}{2} = \frac{5}{EF}$

$EF = 10$ cm

Q4. E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. For $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm, check if $EF \parallel QR$. 3 Marks

Sol. By Converse of BPT, if ratios are equal, lines are parallel.

$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$

$\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{3}{2} = 1.5$

Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, $EF$ is NOT parallel to $QR$.

Q5. In $\triangle ABC$, $DE \parallel BC$ intersecting $AB$ at $D$ and $AC$ at $E$. If $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$, find the value of $x$. 3 Marks (Algebraic BPT)

Sol. Since $DE \parallel BC$, by BPT: $\frac{AD}{DB} = \frac{AE}{EC}$

$$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$$

Cross-multiplying: $x(x - 1) = (x - 2)(x + 2)$

$$x^2 - x = x^2 - 4 \implies -x = -4 \implies \mathbf{x = 4}$$

Check: $AD = 4, DB = 2 \implies 4/2 = 2$; $AE = 6, EC = 3 \implies 6/3 = 2$ ✅ Verified.

🎯 Board Pattern (2018–2025): The proof of the Basic Proportionality Theorem (BPT) is one of the most frequently asked 5-mark questions in geometry. Ensure you write the "Given", "To Prove", and "Construction" parts clearly, as they carry marks even if the proof is incomplete.

📋 Board Revision Checklist

✅ All congruent figures are similar, but similar figures need not be congruent
✅ Two polygons are similar if (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio
✅ BPT Theorem: Parallel line divides other two sides proportionally
✅ AA Similarity: Only 2 angles equal are enough for similarity
✅ SAS Similarity: Ratio of 2 sides and the angle between them are equal
✅ SSS Similarity: Ratios of all three corresponding sides are equal
✅ Order of similarity: $\triangle ABC \sim \triangle PQR$ implies $\angle A=\angle P$, $\angle B=\angle Q$, etc.
✅ Note: Area of Similar Triangles and Pythagoras Theorem are deleted for 2025-26

💡 Exam Tip:
In similarity questions, always name the triangles in the correct order of correspondence. If $\triangle ABC \sim \triangle RQP$, then $\frac{AB}{RQ} = \frac{BC}{QP} = \frac{AC}{RP}$.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

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

2. All ___________ triangles are similar.

3. In $\triangle ABC$, $DE || BC$. If $AD=3$, $DB=5$, $AE=6$, then $EC$ is:

4. If $\triangle ABC \sim \triangle PQR$ such that $\angle A = 45^\circ$ and $\angle B = 75^\circ$, then $\angle R$ is:

5. Two triangles are similar by SAS criterion if:

Chapter 6: Triangles

Overview