Study Guide

Triangles Class 10 CBSE: BPT, Similarity & Pythagoras — Complete Guide

Everything you need to master similar triangles, proportionality theorems, and Pythagoras for your boards!

CBSEClass 10
The SparkEd Authors (IITian & Googler)15 March 202655 min read
Geometric illustration of similar triangles and the Basic Proportionality Theorem for Class 10 CBSE.

Why Triangles Is One of the Highest-Scoring Chapters

If there's one chapter in Class 10 Maths that can make or break your score, it's Triangles. This chapter typically carries 12-15 marks in the CBSE board exam (combined with other geometry chapters), and a solid chunk of that comes directly from similarity and Pythagoras theorem problems.

The good news? Triangles is incredibly systematic. Once you learn the core theorems and the criteria for similarity, you'll find that most problems follow a handful of predictable patterns. Whether it's a BPT application, a similarity proof, or a Pythagoras word problem, the approach is almost always the same: identify the right triangles, set up proportions, and solve.

In this guide, we'll walk through every theorem and concept from NCERT Chapter 6, packed with solved examples, common mistakes to avoid, and a clear strategy for your board exam. Let's get started!

Similar Figures vs. Congruent Figures

Before diving into triangle similarity, let's clarify the difference between similar and congruent figures, because mixing these up is a common source of errors.

Congruent figures have the same shape AND the same size. If you place one on top of the other, they match perfectly. Think of two identical coins.

Similar figures have the same shape but NOT necessarily the same size. One is a scaled version of the other. Think of a photograph and its enlarged print — same proportions, different dimensions.

For polygons, similarity means:
1. All corresponding angles are equal.
2. All corresponding sides are in the same ratio (called the scale factor).

For example, all circles are similar to each other. All squares are similar to each other. But not all rectangles are similar — a 2×32 \times 3 rectangle is not similar to a 2×52 \times 5 rectangle because the ratios of sides differ.

For triangles, something remarkable happens: we don't need to check ALL angles and ALL side ratios. Just a few conditions are enough to establish similarity, which is what makes the similarity criteria so powerful.

Scale Factor and Its Meaning

When two figures are similar, the ratio of their corresponding sides is constant. This constant ratio is called the scale factor (or the ratio of similitude).

If △ABC∼△DEF\triangle ABC \sim \triangle DEF with scale factor kk, then:

ABDE=BCEF=ACDF=k\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k

A scale factor of k=2k = 2 means the first triangle is twice as large as the second. A scale factor of k=13k = \frac{1}{3} means the first triangle is one-third the size of the second.

The order of vertices matters! Writing △ABC∼△DEF\triangle ABC \sim \triangle DEF specifically means A↔DA \leftrightarrow D, B↔EB \leftrightarrow E, C↔FC \leftrightarrow F.

Basic Proportionality Theorem (BPT) — Thales' Theorem

The Basic Proportionality Theorem (BPT), also called Thales' theorem, is the foundation of this entire chapter. It connects parallel lines with proportional division of triangle sides.

Statement (Theorem 6.1): 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.

In △ABC\triangle ABC, if DE∥BCDE \parallel BC where DD lies on ABAB and EE lies on ACAC, then:

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

Why this works (Proof Idea):
Draw DN⊥ACDN \perp AC and EM⊥ABEM \perp AB. Then:

ar(△ADE)=12×AD×EMandar(△BDE)=12×DB×EM\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EM \quad \text{and} \quad \text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EM

∴ar(△ADE)ar(△BDE)=ADDB\therefore \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{AD}{DB}

Similarly, ar(△ADE)ar(△CDE)=AEEC\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{AE}{EC}.

Since △BDE\triangle BDE and △CDE\triangle CDE share the same base DEDE and lie between the same parallels DEDE and BCBC, their areas are equal. This gives us the result.

Equivalent forms of BPT: If DE∥BCDE \parallel BC in △ABC\triangle ABC, then all of these are true:

ADDB=AEEC⟺ADAB=AEAC⟺DBAB=ECAC\frac{AD}{DB} = \frac{AE}{EC} \quad \Longleftrightarrow \quad \frac{AD}{AB} = \frac{AE}{AC} \quad \Longleftrightarrow \quad \frac{DB}{AB} = \frac{EC}{AC}

Converse of BPT (Theorem 6.2)

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

In △ABC\triangle ABC, if DD lies on ABAB and EE lies on ACAC such that ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}, then DE∥BCDE \parallel BC.

The converse is extremely useful in problems where you need to prove that two lines are parallel. You simply compute the ratios and show they're equal.

Solved Example 1: Direct BPT Application

Problem: In △ABC\triangle ABC, DE∥BCDE \parallel BC with DD on ABAB and EE on ACAC. If AD=4AD = 4 cm, DB=6DB = 6 cm, and AE=3AE = 3 cm, find ECEC.

Solution:
By BPT, since DE∥BCDE \parallel BC:

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

46=3EC\frac{4}{6} = \frac{3}{EC}

EC=3×64=184=4.5 cmEC = \frac{3 \times 6}{4} = \frac{18}{4} = 4.5 \text{ cm}

Answer: EC=4.5EC = 4.5 cm.

Solved Example 2: Converse of BPT

Problem: In △PQR\triangle PQR, SS lies on PQPQ and TT lies on PRPR such that PS=3PS = 3 cm, SQ=5SQ = 5 cm, PT=4.5PT = 4.5 cm, and TR=7.5TR = 7.5 cm. Is ST∥QRST \parallel QR?

Solution:

PSSQ=35=0.6\frac{PS}{SQ} = \frac{3}{5} = 0.6

PTTR=4.57.5=35=0.6\frac{PT}{TR} = \frac{4.5}{7.5} = \frac{3}{5} = 0.6

Since PSSQ=PTTR\frac{PS}{SQ} = \frac{PT}{TR}, by the converse of BPT, ST∥QRST \parallel QR. □\square

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Criteria for Similarity of Triangles

Now we come to the heart of this chapter — the three criteria that let us prove two triangles are similar without checking every angle and every side ratio.

Remember: for similarity, we write △ABC∼△DEF\triangle ABC \sim \triangle DEF, meaning corresponding angles are equal and corresponding sides are proportional.

AA (Angle-Angle) Similarity Criterion — Theorem 6.3

Statement: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (i.e., they are similar).

If ∠A=∠D\angle A = \angle D and ∠B=∠E\angle B = \angle E, then △ABC∼△DEF\triangle ABC \sim \triangle DEF.

Why only two angles? Since the sum of angles in a triangle is always 180∘180^\circ, if two angles of one triangle equal two angles of another, the third angles are automatically equal too.

This is the most commonly used criterion in board exams. Whenever you spot two pairs of equal angles, you can immediately conclude similarity.

SSS (Side-Side-Side) Similarity Criterion — Theorem 6.4

Statement: If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then the two triangles are similar.

If ABDE=BCEF=ACDF\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}, then △ABC∼△DEF\triangle ABC \sim \triangle DEF.

Example: If △ABC\triangle ABC has sides 3,4,53, 4, 5 and △DEF\triangle DEF has sides 6,8,106, 8, 10, then:

36=48=510=12\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}

All ratios are equal, so △ABC∼△DEF\triangle ABC \sim \triangle DEF by SSS similarity, with scale factor 12\frac{1}{2}.

SAS (Side-Angle-Side) Similarity Criterion — Theorem 6.5

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

If ∠A=∠D\angle A = \angle D and ABDE=ACDF\frac{AB}{DE} = \frac{AC}{DF}, then △ABC∼△DEF\triangle ABC \sim \triangle DEF.

Important: The equal angle must be the included angle (the angle between the two proportional sides). This is a common mistake — students sometimes use a non-included angle and still claim SAS similarity.

Solved Example 3: Using AA Similarity

Problem: In the figure, ∠AEP=∠CDP=90∘\angle AEP = \angle CDP = 90^\circ. Prove that △AEP∼△CDP\triangle AEP \sim \triangle CDP.

Solution:
In △AEP\triangle AEP and △CDP\triangle CDP:
1. ∠AEP=∠CDP=90∘\angle AEP = \angle CDP = 90^\circ (Given)
2. ∠APE=∠CPD\angle APE = \angle CPD (Vertically opposite angles)

Since two pairs of corresponding angles are equal, by the AA similarity criterion:

△AEP∼△CDP□\triangle AEP \sim \triangle CDP \quad \square

Areas of Similar Triangles — Theorem 6.6

This theorem connects the areas of similar triangles to the ratio of their corresponding sides, and it appears frequently in board exams.

Statement (Theorem 6.6): The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If △ABC∼△DEF\triangle ABC \sim \triangle DEF, then:

ar(△ABC)ar(△DEF)=(ABDE)2=(BCEF)2=(ACDF)2\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2

Why the square? Area is a two-dimensional measure. When you scale a figure by factor kk, each linear dimension scales by kk, but area scales by k2k^2.

Corollaries that follow:
- The ratio of areas also equals the square of the ratio of corresponding altitudes.
- The ratio of areas equals the square of the ratio of corresponding medians.
- The ratio of areas equals the square of the ratio of corresponding angle bisectors.
- The ratio of perimeters of similar triangles equals the ratio of corresponding sides (not squared).

Solved Example 4: Areas of Similar Triangles

Problem: If △ABC∼△DEF\triangle ABC \sim \triangle DEF such that AB=5AB = 5 cm, DE=8DE = 8 cm, and ar(△ABC)=50\text{ar}(\triangle ABC) = 50 sq cm, find ar(△DEF)\text{ar}(\triangle DEF).

Solution:
Using Theorem 6.6:

ar(△ABC)ar(△DEF)=(ABDE)2\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2

50ar(△DEF)=(58)2=2564\frac{50}{\text{ar}(\triangle DEF)} = \left(\frac{5}{8}\right)^2 = \frac{25}{64}

ar(△DEF)=50×6425=320025=128 sq cm\text{ar}(\triangle DEF) = \frac{50 \times 64}{25} = \frac{3200}{25} = 128 \text{ sq cm}

Answer: ar(△DEF)=128\text{ar}(\triangle DEF) = 128 sq cm.

Solved Example 5: Finding Side from Area Ratio

Problem: The areas of two similar triangles are 8181 sq cm and 4949 sq cm. If the altitude of the larger triangle is 6.36.3 cm, find the corresponding altitude of the smaller triangle.

Solution:
Let the corresponding altitude of the smaller triangle be hh.

Since the ratio of areas equals the square of the ratio of corresponding altitudes:

8149=(6.3h)2\frac{81}{49} = \left(\frac{6.3}{h}\right)^2

6.3h=8149=97\frac{6.3}{h} = \sqrt{\frac{81}{49}} = \frac{9}{7}

h=6.3×79=44.19=4.9 cmh = \frac{6.3 \times 7}{9} = \frac{44.1}{9} = 4.9 \text{ cm}

Answer: The altitude of the smaller triangle is 4.94.9 cm.

Pythagoras Theorem — Theorem 6.8

The most famous theorem in all of mathematics, and a guaranteed question in your board exam.

Statement (Theorem 6.8): In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If △ABC\triangle ABC is right-angled at BB, then:

AC2=AB2+BC2AC^2 = AB^2 + BC^2

Proof using Similarity (NCERT approach):
In △ABC\triangle ABC with ∠B=90∘\angle B = 90^\circ, draw altitude BD⊥ACBD \perp AC.

This creates three similar triangles:
- △ADB∼△ABC\triangle ADB \sim \triangle ABC (AA: ∠A\angle A is common, both have a right angle)
- △BDC∼△ABC\triangle BDC \sim \triangle ABC (AA: ∠C\angle C is common, both have a right angle)

From △ADB∼△ABC\triangle ADB \sim \triangle ABC:

ADAB=ABAC  ⟹  AD⋅AC=AB2⋯(1)\frac{AD}{AB} = \frac{AB}{AC} \implies AD \cdot AC = AB^2 \quad \cdots (1)

From △BDC∼△ABC\triangle BDC \sim \triangle ABC:

CDBC=BCAC  ⟹  CD⋅AC=BC2⋯(2)\frac{CD}{BC} = \frac{BC}{AC} \implies CD \cdot AC = BC^2 \quad \cdots (2)

Adding (1) and (2):

AD⋅AC+CD⋅AC=AB2+BC2AD \cdot AC + CD \cdot AC = AB^2 + BC^2

AC(AD+CD)=AB2+BC2AC(AD + CD) = AB^2 + BC^2

AC⋅AC=AB2+BC2AC \cdot AC = AB^2 + BC^2

AC2=AB2+BC2□AC^2 = AB^2 + BC^2 \quad \square

Converse of Pythagoras Theorem — Theorem 6.9

Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

If AC2=AB2+BC2AC^2 = AB^2 + BC^2 in △ABC\triangle ABC, then ∠B=90∘\angle B = 90^\circ.

This converse is used to verify whether a triangle is right-angled when you know all three sides.

Solved Example 6: Pythagoras Theorem

Problem: A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution:
Let the wall be AB=8AB = 8 m, ladder be AC=10AC = 10 m, and ground distance be BCBC.

Since the wall is perpendicular to the ground, ∠B=90∘\angle B = 90^\circ.

By Pythagoras theorem:

AC2=AB2+BC2AC^2 = AB^2 + BC^2

102=82+BC210^2 = 8^2 + BC^2

100=64+BC2100 = 64 + BC^2

BC2=36BC^2 = 36

BC=6 mBC = 6 \text{ m}

Answer: The foot of the ladder is 66 m from the wall.

Solved Example 7: Converse of Pythagoras

Problem: A triangle has sides 77 cm, 2424 cm, and 2525 cm. Is it a right triangle?

Solution:
The longest side is 2525 cm. Check if 252=72+24225^2 = 7^2 + 24^2:

252=62525^2 = 625

72+242=49+576=6257^2 + 24^2 = 49 + 576 = 625

Since 252=72+24225^2 = 7^2 + 24^2, by the converse of Pythagoras theorem, the triangle is right-angled, with the right angle opposite the side of 2525 cm. □\square

Important Results Built on Pythagoras Theorem

Several important results in the NCERT chapter are direct applications of the Pythagoras theorem. These appear regularly in board exams and you should know them well.

Altitude on Hypotenuse

If an altitude is drawn from the right-angle vertex to the hypotenuse of a right triangle, the two triangles formed are similar to each other and to the original triangle.

In △ABC\triangle ABC with ∠B=90∘\angle B = 90^\circ and BD⊥ACBD \perp AC:
- △ADB∼△ABC∼△BDC\triangle ADB \sim \triangle ABC \sim \triangle BDC
- BD2=AD×DCBD^2 = AD \times DC (geometric mean relation)
- AB2=AD×ACAB^2 = AD \times AC
- BC2=CD×CABC^2 = CD \times CA

These results are extremely useful for solving problems involving altitudes in right triangles.

Solved Example 8: Altitude on Hypotenuse

Problem: In △ABC\triangle ABC, ∠B=90∘\angle B = 90^\circ, and BD⊥ACBD \perp AC. If AD=3AD = 3 cm and DC=12DC = 12 cm, find BDBD.

Solution:
Using the geometric mean relation:

BD2=AD×DC=3×12=36BD^2 = AD \times DC = 3 \times 12 = 36

BD=6 cmBD = 6 \text{ cm}

Answer: BD=6BD = 6 cm.

In an Equilateral Triangle

Result: In an equilateral triangle with side aa, the altitude hh is given by:

h=32ah = \frac{\sqrt{3}}{2} a

Proof: The altitude of an equilateral triangle bisects the base. If the side is aa, then by Pythagoras theorem in the right triangle formed:

a2=h2+(a2)2a^2 = h^2 + \left(\frac{a}{2}\right)^2

h2=a2−a24=3a24h^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}

h=32ah = \frac{\sqrt{3}}{2} a

Area of equilateral triangle:

Area=12×a×32a=34a2\text{Area} = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{4} a^2

More Solved Examples (Board Exam Level)

Here are additional solved examples that match the difficulty level and style of CBSE board exam questions.

Solved Example 9: BPT with Midpoint

Problem: In △ABC\triangle ABC, DD is the midpoint of ABAB and DE∥BCDE \parallel BC where EE lies on ACAC. Prove that EE is the midpoint of ACAC.

Solution:
Since DD is the midpoint of ABAB, we have AD=DBAD = DB, so ADDB=1\frac{AD}{DB} = 1.

By BPT, since DE∥BCDE \parallel BC:

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

1=AEEC1 = \frac{AE}{EC}

AE=ECAE = EC

Therefore EE is the midpoint of ACAC. □\square

Note: This is actually the proof of the Midpoint Theorem using BPT!

Solved Example 10: Combining Similarity with Area

Problem: In △ABC\triangle ABC, PQ∥BCPQ \parallel BC with PP on ABAB and QQ on ACAC. If APPB=23\frac{AP}{PB} = \frac{2}{3} and ar(△APQ)=32\text{ar}(\triangle APQ) = 32 sq cm, find ar(△ABC)\text{ar}(\triangle ABC).

Solution:
Since PQ∥BCPQ \parallel BC, by AA similarity: △APQ∼△ABC\triangle APQ \sim \triangle ABC.

Given APPB=23\frac{AP}{PB} = \frac{2}{3}, so APAB=APAP+PB=22+3=25\frac{AP}{AB} = \frac{AP}{AP + PB} = \frac{2}{2+3} = \frac{2}{5}.

By Theorem 6.6:

ar(△APQ)ar(△ABC)=(APAB)2=(25)2=425\frac{\text{ar}(\triangle APQ)}{\text{ar}(\triangle ABC)} = \left(\frac{AP}{AB}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}

32ar(△ABC)=425\frac{32}{\text{ar}(\triangle ABC)} = \frac{4}{25}

ar(△ABC)=32×254=200 sq cm\text{ar}(\triangle ABC) = \frac{32 \times 25}{4} = 200 \text{ sq cm}

Answer: ar(△ABC)=200\text{ar}(\triangle ABC) = 200 sq cm.

Solved Example 11: Proving a Right Angle

Problem: In a quadrilateral ABCDABCD, ∠B=90∘\angle B = 90^\circ. If AD2=AB2+BC2+CD2AD^2 = AB^2 + BC^2 + CD^2, prove that ∠ACD=90∘\angle ACD = 90^\circ.

Solution:
In △ABC\triangle ABC with ∠B=90∘\angle B = 90^\circ, by Pythagoras theorem:

AC2=AB2+BC2⋯(1)AC^2 = AB^2 + BC^2 \quad \cdots (1)

Given: AD2=AB2+BC2+CD2AD^2 = AB^2 + BC^2 + CD^2

Substituting from (1):

AD2=AC2+CD2AD^2 = AC^2 + CD^2

By the converse of Pythagoras theorem in △ACD\triangle ACD:

∠ACD=90∘□\angle ACD = 90^\circ \quad \square

Common Mistakes Students Make in Triangles

Here are the most frequent errors that cost students marks in board exams — learn these and you'll be ahead of the curve:

1. Wrong Order of Vertices in Similarity:
* Mistake: Writing △ABC∼△EDF\triangle ABC \sim \triangle EDF when the correct correspondence is A↔DA \leftrightarrow D, B↔EB \leftrightarrow E, C↔FC \leftrightarrow F.
* Fix: The order of letters indicates which vertices correspond. Always match corresponding angles first, then write the similarity statement accordingly.

2. Confusing Similarity with Congruence:
* Mistake: Claiming that similar triangles have equal sides.
* Fix: Similar means proportional sides and equal angles. Congruent means equal sides and equal angles. Similar triangles are generally different sizes.

3. Using Non-Included Angle in SAS Similarity:
* Mistake: Claiming SAS similarity when the equal angle is not the angle between the two proportional sides.
* Fix: In SAS similarity, the equal angle must be the included angle between the two sides whose ratios are equal.

4. Forgetting to Check All Three Ratios in SSS:
* Mistake: Checking only two ratios and assuming the third is also equal.
* Fix: For SSS similarity, verify that all three ratios of corresponding sides are equal.

5. Area Ratio vs. Side Ratio:
* Mistake: Using the ratio of sides directly as the ratio of areas.
* Fix: The area ratio is the square of the side ratio. If sides are in ratio 3:53:5, areas are in ratio 9:259:25, not 3:53:5.

6. BPT Applied Incorrectly:
* Mistake: Using BPT when the line is NOT parallel to one side of the triangle.
* Fix: BPT only applies when the line is parallel to one side. If the problem doesn't explicitly state parallelism, you cannot assume it.

7. Not Drawing or Labelling Diagrams:
* Mistake: Attempting problems mentally without a clear diagram.
* Fix: Always draw a diagram. Label all given lengths, angles, and mark the parallel lines or right angles. A good diagram is half the solution.

Board Exam Strategy: Weightage and Question Patterns

The Triangles chapter is one of the most important chapters in CBSE Class 10 Maths. Here's what you need to know for exam day:

Weightage: The chapter typically carries 8-10 marks directly (from Unit IV: Geometry, which includes Triangles and Circles). Questions from this chapter appear across all question types.

Typical Question Patterns:

* 1 Mark (MCQ/VSA): Direct application of BPT with simple numbers; checking if a triangle is right-angled using the converse of Pythagoras; identifying corresponding sides in similar triangles.

* 2-3 Marks (SA): Finding unknown sides using BPT; proving two triangles are similar using AA/SAS/SSS; finding area ratio of similar triangles; simple Pythagoras word problems.

* 4-5 Marks (LA): Proof of BPT (Theorem 6.1); proof of Pythagoras theorem using similarity; multi-step problems combining similarity with areas; problems involving altitudes on the hypotenuse.

High-Priority Topics for Board Exams:
1. Proof of BPT (Theorem 6.1) — asked very frequently
2. Proof of Pythagoras theorem (Theorem 6.8) — almost guaranteed
3. AA similarity problems with numerical calculations
4. Area ratio problems using Theorem 6.6
5. Problems combining BPT with midpoint theorem

Time Allocation: For a 5-mark proof question, spend about 6-8 minutes. For a 3-mark numerical, spend about 4-5 minutes. Always draw diagrams and state reasons clearly.

Practice on SparkEd's Triangles topic page to get a feel for various difficulty levels.

Quick Reference: All Theorems at a Glance

Here's your revision cheat sheet — bookmark this section for quick review before exams:

Theorem 6.1 (BPT): DE∥BCDE \parallel BC in △ABC  ⟹  ADDB=AEEC\triangle ABC \implies \frac{AD}{DB} = \frac{AE}{EC}

Theorem 6.2 (Converse of BPT): ADDB=AEEC  ⟹  DE∥BC\frac{AD}{DB} = \frac{AE}{EC} \implies DE \parallel BC

Theorem 6.3 (AA Similarity): Two angles equal   ⟹  \implies triangles similar

Theorem 6.4 (SSS Similarity): All three side ratios equal   ⟹  \implies triangles similar

Theorem 6.5 (SAS Similarity): One angle equal + including sides proportional   ⟹  \implies triangles similar

Theorem 6.6 (Area Ratio): ar(△1)ar(△2)=(side1side2)2\frac{\text{ar}(\triangle_1)}{\text{ar}(\triangle_2)} = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2

Theorem 6.8 (Pythagoras): In a right △\triangle, hypotenuse2=base2+height2\text{hypotenuse}^2 = \text{base}^2 + \text{height}^2

Theorem 6.9 (Converse of Pythagoras): If c2=a2+b2c^2 = a^2 + b^2, then ∠\angle opposite to side cc is 90∘90^\circ

Key Formulas:
- Equilateral triangle altitude: h=32ah = \frac{\sqrt{3}}{2}a
- Equilateral triangle area: 34a2\frac{\sqrt{3}}{4}a^2
- Geometric mean relation: BD2=AD×DCBD^2 = AD \times DC (altitude on hypotenuse)

Boost Your Preparation with SparkEd

You've just gone through the entire Triangles chapter — every theorem, every criterion, every problem type. But reading alone won't get you full marks; practice will.

Here's how SparkEd can help you ace Triangles and every other chapter:

* Practice by Difficulty: On our Triangles practice page, work through problems sorted into Level 1, Level 2, and Level 3. Build confidence gradually.

* AI Math Solver: Stuck on a BPT proof or a tricky similarity problem? Paste it into our AI Solver and get step-by-step solutions with detailed reasoning.

* AI Coach: Get personalized recommendations on which topics need more practice based on your performance. The Coach spots your weak areas before the examiner does.

* Cross-Topic Connections: Triangles connects to Coordinate Geometry (distance formula uses Pythagoras!), Trigonometry (right-triangle ratios), and Circles (tangent-radius problems). Explore all of these on our programs page.

Head over to sparkedmaths.com and start practicing today. Every problem you solve now is a mark earned on exam day!

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