Overview
Chapter 'Triangles' (Class 9, NCERT) introduces triangles, their types and basic elements, and develops students' understanding of triangle congruence, key properties and inequalities. The chapter is important because triangles are a fundamental geometric shape used throughout geometry and in many applications; mastering congruence and basic theorems builds proof skills and logical reasoning. Key themes include definitions (vertices, sides, angles), congruence criteria (SSS, SAS, ASA, RHS), properties of isosceles and general triangles, and triangle inequalities (relation between sides and opposite angles). By the end of the chapter students will be able to identify and classify triangles, state and use congruence criteria to prove two triangles are congruent, apply basic theorems (e.g., base angles of an isosceles triangle are equal and converses), use inequalities relating sides and angles, and solve construction and problem-solving exercises that reinforce geometric proof and reasoning.
Learning Objectives
- Define triangle and classify triangles by sides (scalene, isosceles, equilateral) and by angles (acute, right, obtuse).
- State and apply SSS, SAS, ASA and RHS criteria to test congruence of two triangles in exam problems.
- Prove congruence results and use congruence to determine unknown sides and angles in given figures.
- Explain and apply the angle-sum property of a triangle (sum of interior angles = 180°) to solve numerical and proof-based questions.
- Derive and use the Exterior Angle Theorem to calculate angles and justify steps in proofs.
- Prove that base angles of an isosceles triangle are equal and apply this result to solve related problems.
- State and use the relationship between sides and opposite angles (a larger side is opposite a larger angle) to compare elements of a triangle.
- Apply the triangle inequality theorem to determine feasible side lengths and to verify whether three lengths form a triangle.
Topics in this chapter
8 topics · tap a topic title to jump straight to it.
Basic concepts and terminology
A triangle is a polygon with three sides and three angles formed by joining three non-collinear points. If the vertices are A, B and C, the triangle is denoted by ABC. Basic concepts and terminology help us describe, classify and work with triangles.
Key terms
- Vertices: The three corner points (A, B, C).
- Sides: The line segments AB, BC and CA.
- Angles: ∠A, ∠B, ∠C are the interior angles at the vertices.
- Base: Any side chosen as reference (often BC when talking about altitude from A).
- Altitude (height): A perpendicular from a vertex to the opposite side (or its extension).
- Median: A line segment joining a vertex to the midpoint of the opposite side.
- Angle bisector: A ray from a vertex that divides the angle into two equal angles.
- Perpendicular bisector: A line perpendicular to a side at its midpoint.
- Interior and exterior angles: Interior are inside the triangle; an exterior angle is formed by extending a side.
Classification by sides
- Scalene: All three sides (and all three angles) are different.
- Isosceles: Two sides (and two angles) are equal.
- Equilateral: All three sides and angles are equal (each angle = 60°).
Classification by angles
- Acute: All three interior angles are less than 90°.
- Right: One interior angle is exactly 90°.
- Obtuse: One interior angle is greater than 90°.
Important basic properties
- The sum of the interior angles: ∠A + ∠B + ∠C = 180°.
- Exterior angle property: An exterior angle equals the sum of the two opposite interior angles (e.g., exterior at A = ∠B + ∠C).
- Triangle inequality: The sum of lengths of any two sides is greater than the third side (AB + BC > CA, etc.).
- Perimeter = AB + BC + CA.
- Area (basic): Area = 1/2 × base × height (use the altitude to the chosen base).
- Concurrency of special lines: medians meet at the centroid (which divides each median in 2:1 ratio); angle bisectors meet at the incenter; perpendicular bisectors meet at the circumcenter; altitudes meet at the orthocenter.
These definitions and properties form the foundation for studying congruence, similarity, constructions and many real-world applications such as surveying, engineering and architecture.
- Roof trusses: Triangular frames are used because triangles are rigid shapes—three bars joined at three joints form a stable structure.
- Traffic signs: Many road signs are triangular (e.g., yield sign) making them easily recognizable from shape.
- Surveying and navigation: Triangulation uses triangles and angle measurements to determine distances and positions.
- Bridge design: Triangular components (trusses) distribute loads efficiently and prevent bending.
- Sails and masts: Triangular sails and stays provide strength and control wind forces on boats.
- Sum of interior angles: ∠A + ∠B + ∠C = 180°
- Exterior angle: Exterior at A = ∠B + ∠C
- Triangle inequality: AB + BC > CA, BC + CA > AB, CA + AB > BC
- Perimeter: P = AB + BC + CA
- Area (base and height): Area = (1/2) × base × height
- Right triangle (Pythagoras): If ∠C = 90°, then AB^2 = AC^2 + BC^2
Triangle inequalities and side–angle relations
Triangle inequality (statement): In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. If the sides are a, b, c then
a + b > c, b + c > a, c + a > b.
Proof (short intuitive): In triangle ABC, to go from B to C directly is shorter than going B → A → C. So BC < BA + AC. The same reasoning gives the other two inequalities.
Related inequality (difference): The third side is also greater than the positive difference of the other two sides:
|a - b| < c < a + b.
Equality a + b = c means the three points are collinear (no triangle, they lie on a straight line).
Side–angle relations (statement):
- If one angle of a triangle is greater than another angle, then the side opposite the greater angle is longer than the side opposite the smaller angle. (If ∠A > ∠B then side a > side b.)
- Converse: If one side is longer than another side, then the angle opposite the longer side is larger. (If a > b then ∠A > ∠B.)
Sketch of proof for side–angle relation: In triangle ABC assume ∠A > ∠B. Mark a point D on AC so that AD = AB. Then triangle ADB is isosceles so ∠ADB = ∠B. Because ∠A > ∠B, the remaining angle ∠ADC (or angle at D looking toward C) is positive so DC > 0 hence AC > AB, that is side opposite the larger angle is longer. The converse follows by reversing the argument.
Useful consequences:
- To test if three segments can form a triangle check the triangle inequalities.
- The largest side is opposite the largest angle; the smallest side opposite the smallest angle.
- Exterior angle theorem: an exterior angle is greater than either remote interior angle; this supports side–angle comparisons.
Applications: checks for constructibility of triangles from given sticks, designing stable tripods and trusses, deciding which roof rafter or beam must be longer given angles, and quick feasibility tests in geometry problems.
- Feasibility check: Can lengths 3, 4, 8 form a triangle? Check 3 + 4 = 7 which is not greater than 8, so no triangle can be formed.
- Valid triangle: Lengths 3, 4, 6. Check 3 + 4 = 7 > 6, 4 + 6 = 10 > 3, 6 + 3 = 9 > 4. So a triangle is possible. The largest side 6 is opposite the largest angle.
- Side–angle relation: In triangle ABC, if sides are a = 7, b = 5, c = 6, then a is largest so angle A is largest (∠A > ∠C > ∠B).
- Practical: Three wooden beams of lengths 2 m, 2.5 m, 5 m cannot form a triangular frame because 2 + 2.5 = 4.5 < 5.
- Triangle inequality: a + b > c, b + c > a, c + a > b
- Difference form: |a - b| < c < a + b
- No triangle (collinear) case: a + b = c (points lie on a straight line)
- Side–angle relation: If ∠A > ∠B then a > b. Converse: If a > b then ∠A > ∠B
- Exterior angle: An exterior angle > either remote interior angle (supports side–angle comparisons)
Congruence of triangles
Definition: Two triangles are congruent if all corresponding sides and corresponding angles are equal. Congruence is denoted by the symbol ≅ (for example, ΔABC ≅ ΔDEF means AB = DE, BC = EF, CA = FD and ∠A = ∠D, ∠B = ∠E, ∠C = ∠F).
Correspondence: To claim congruence you must pair vertices so that corresponding sides and angles match. The order in the notation ΔABC ≅ ΔDEF indicates which vertices correspond: A↔D, B↔E, C↔F.
Congruence criteria (Class 9 focus):
- SSS (Side–Side–Side): If three sides of one triangle are equal to three sides of another, the triangles are congruent.
- SAS (Side–Included Angle–Side): If two sides and the included angle of one triangle equal the corresponding two sides and included angle of another, the triangles are congruent.
- ASA (Angle–Side–Angle): If two angles and the included side of one triangle equal the corresponding two angles and included side of another, the triangles are congruent.
- RHS (Right angle–Hypotenuse–Side): For right triangles, if the hypotenuse and one corresponding leg of two triangles are equal, the triangles are congruent.
Why these work (brief): Each criterion fixes enough information so that the third side/angle is uniquely determined (no other non-congruent triangle can be formed). Note: SSA (two sides and a non-included angle) does not generally guarantee congruence.
Consequences: Once triangles are proved congruent, corresponding parts are equal (CPCTC — Corresponding Parts of Congruent Triangles are Congruent). Congruence can also be understood as one triangle being obtainable from the other by rigid motions (translation, rotation, reflection) without resizing.
Practical tip: Always specify the matching order of vertices when stating congruence and state which criterion you use.
- Carpentry: When cutting identical triangular pieces to form a frame or truss, workers use congruence (same sides and angles) to ensure pieces fit exactly.
- Bridge/roof trusses: Engineers design repeating congruent triangular units so loads and stresses are uniformly distributed.
- Manufacturing: Identical triangular metal plates or brackets are produced by ensuring dimensions match (SSS or SAS) so they are interchangeable.
- Tiling and pavements: Using congruent equilateral or isosceles triangles for aesthetic patterns guarantees perfect fits without gaps.
- Surveying and mapping: To verify that two triangular parcels are identical in shape and size, surveyors check corresponding side lengths and angles (coordinate method or distance checks).
- If ΔABC ≅ ΔDEF then AB = DE, BC = EF, CA = FD and ∠A = ∠D, ∠B = ∠E, ∠C = ∠F (CPCTC).
- SSS criterion: AB = DE, BC = EF, CA = FD ⇒ ΔABC ≅ ΔDEF.
- SAS criterion: AB = DE, ∠B = ∠E (included), BC = EF ⇒ ΔABC ≅ ΔDEF.
- ASA criterion: ∠A = ∠D, AB = DE (included), ∠B = ∠E ⇒ ΔABC ≅ ΔDEF.
- RHS criterion (right triangles): hypotenuse and one leg equal ⇒ triangles congruent.
- Distance formula (useful in coordinate geometry): length between (x1,y1) and (x2,y2) = √((x2−x1)^2 + (y2−y1)^2). Use this to verify equal sides when coordinates are given.
Properties of special triangles
What are special triangles? Special triangles are triangles with particular side or angle relations that give them useful and easily remembered properties. The main special types studied in Class 9 are equilateral, isosceles and right-angled triangles (including the commonly used 45°-45°-90° and 30°-60°-90° right triangles).
- Equilateral triangle (all sides equal):
- All three sides are equal: a = b = c.
- All interior angles are equal and each = 60°.
- The medians, altitudes, perpendicular bisectors and angle bisectors from any vertex coincide (a single line does all these jobs).
- Centroid, circumcenter, incenter and orthocenter are the same point.
- Isosceles triangle (two sides equal):
- If two sides are equal (say AB = AC), the angles opposite those sides are equal (∠B = ∠C). This follows by triangle congruence (S-S-S or S-A-S arguments).
- The line from the vertex between the equal sides to the midpoint of the base is simultaneously the median, altitude and angle bisector.
- Two equal sides make the triangle symmetric about the altitude from the vertex between them.
- Right-angled triangle:
- One angle is 90°. The side opposite it is the hypotenuse (longest side).
- Pythagorean theorem: for legs a and b and hypotenuse c, a² + b² = c².
- Altitude from the right angle to the hypotenuse creates two smaller right triangles, each similar to the original (useful for many constructions and proofs).
- Special right triangles:
- 45°-45°-90°: legs equal; hypotenuse = side × √2.
- 30°-60°-90°: if hypotenuse = 2k, shorter leg (opposite 30°) = k and longer leg = k√3.
- Notable points and general facts (apply to many triangles):
- Centroid (G): intersection of medians; it divides each median in the ratio 2:1 (vertex to centroid : centroid to midpoint = 2:1).
- Incenter: intersection of angle bisectors; center of the inscribed circle (equidistant from all sides).
- Circumcenter: intersection of perpendicular bisectors of sides; center of circumcircle (equidistant from vertices). In a right triangle the circumcenter is at the midpoint of the hypotenuse.
- Orthocenter: intersection of altitudes. In an equilateral triangle all four centers coincide.
Short proofs/ideas:
- Isosceles base angles equal: Draw equal sides AB = AC. Triangles ABC and ACB are congruent by S-S-S (or S-A-S), so ∠B = ∠C.
- Equilateral angles 60°: From a = b = c, base angles equal and sum of angles 180°, so each = 60°.
- Pythagoras (idea): For a right triangle with legs a, b and hypotenuse c, squares on legs rearrange to fill square on hypotenuse (standard geometrical proof) leading to a² + b² = c².
- Equilateral: Triangular tiles or equilateral-triangle-based patterns in floor design and some honeycomb approximations — each small triangle has all sides equal, so angles are 60° and centers coincide.
- Isosceles: A typical pitched-roof house or an isosceles tent: the two sloping sides are equal, so the roof looks symmetric and the altitude from the peak bisects the base.
- Right triangle: A staircase or a ramp forms a right-angled triangle with vertical rise and horizontal run; Pythagoras helps compute the length of the slope.
- 45°-45°-90°: A square cut along its diagonal produces two congruent 45°-45°-90° triangles. Hypotenuse = side × √2.
- 30°-60°-90°: Splitting an equilateral triangle by an altitude gives two 30°-60°-90° triangles; useful to find heights and divide areas.
- Equilateral triangle of side s: height h = (√3 / 2) s.
- Equilateral area: A = (√3 / 4) s².
- Isosceles triangle with equal sides a and base b: altitude from apex h = √(a² - (b² / 4)).
- Right triangle (legs a, b; hypotenuse c): Pythagorean theorem a² + b² = c².
- Area of right triangle: A = (1/2) × (leg1) × (leg2) = (1/2)ab.
- 45°-45°-90°: if leg = x then hypotenuse = x√2.
Similarity of triangles
Definition: Two triangles are said to be similar if their corresponding angles are equal and the lengths of their corresponding sides are in proportion. In symbols, if ΔABC ~ ΔDEF then ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD.
Why it matters: Similarity tells us shape is the same though size may differ. It lets us find unknown lengths, heights, distances, and is widely used in measurements, maps and models.
Basic criteria (tests) for similarity:
- AA (Angle-Angle): If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. (Because the third angles must also be equal.)
- SAS (Side-Angle-Side) similarity: If two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, the triangles are similar.
- SSS (Side-Side-Side) similarity: If the three sides of one triangle are proportional to the three sides of another triangle, the triangles are similar.
Consequences / properties:
- Corresponding angles are equal and corresponding sides are in the same ratio (called the similarity ratio or scale factor k).
- Ratio of perimeters = k (same as ratio of corresponding sides).
- Ratio of areas = k² (square of similarity ratio).
- Corresponding altitudes, medians and angle bisectors are also in the ratio k.
- If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally and creates a smaller triangle similar to the whole (Basic Proportionality Theorem / Thales' theorem).
How to use similarity in problems:
- Identify corresponding angles or parallel lines to establish equality of angles (AA).
- Set up proportionality equations for corresponding sides to find unknown lengths.
- Use area relations when areas are involved: Area ratio = (side ratio)².
Short proof idea for AA criterion: If two angles of ΔABC equal two angles of ΔDEF, then the third angles match too (sum of angles 180°). Thus all three angles correspond; triangles with equal corresponding angles are similar because corresponding sides opposite equal angles are in proportion (this can be shown using parallel lines or by constructing equal angles and using congruence arguments).
- Shadow method: A pole and its shadow form a right-angled triangle; a person and their shadow form a similar triangle. Using lengths of shadows we can find height of the pole (height/pole shadow = person height/person shadow).
- Scale models and maps: A map uses a constant scale factor to represent distances; triangles formed by landmarks on map are similar to triangles formed on ground.
- Architecture and engineering: Similar triangles help design scaled versions of a structure and compute heights using sight-lines from a point.
- Camera perspective: Objects farther away appear smaller; corresponding triangles from camera lens to object form similar triangles, useful in photogrammetry.
- Using a ladder against a wall: Triangles formed by ladder, wall and ground for different ladder positions are similar if the angle with the ground is same; can compute reach or height.
- If ΔABC ~ ΔDEF then AB/DE = BC/EF = CA/FD (corresponding sides are proportional).
- Similarity ratio (scale factor) k = AB/DE = BC/EF = CA/FD.
- Perimeter ratio = k (Perimeter(ΔABC)/Perimeter(ΔDEF) = k).
- Area ratio = k² (Area(ΔABC)/Area(ΔDEF) = k^2).
- AA criterion: If ∠A = ∠D and ∠B = ∠E then ΔABC ~ ΔDEF.
- SAS similarity: If AB/DE = AC/DF and ∠A = ∠D then ΔABC ~ ΔDEF.
Basic Proportionality Theorem (Thales) and converse
Statement (Basic Proportionality Theorem / Thales): If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio.
More concretely, in triangle ABC, let a line through D on AB and E on AC be parallel to BC (so DE || BC). Then
AD/DB = AE/EC
Proof (using similar triangles):
- Since DE || BC, angle ADE = angle ABC (corresponding angles) and angle AEC = angle ACB.
- Thus triangle ADE is similar to triangle ABC (AA similarity).
- From similarity, corresponding sides are proportional: AD/AB = AE/AC = DE/BC.
- From AD/AB = AE/AC we get AD/(AD+DB) = AE/(AE+EC). Rearranging gives AD/DB = AE/EC, which is the required result.
Converse: If a line through points D on AB and E on AC satisfies AD/DB = AE/EC, then the line DE is parallel to BC.
Proof idea (construction method): Assume AD/DB = AE/EC. Construct a point E' on AC so that DE' is parallel to BC. By the Basic Proportionality Theorem applied to DE' || BC we have AD/DB = AE'/E'C. By the given equality and uniqueness of partition points on AC, AE' = AE, so E' = E. Hence DE || BC.
Useful corollaries and extensions:
- If several parallel lines intersect two sides of a triangle, they divide those sides proportionally (all corresponding segments have equal ratios).
- AD/AB = AE/AC = DE/BC (the common ratio from triangle similarity).
How to use it: The theorem is often used to find unknown segment lengths in triangles by setting up proportional relations and solving linear equations.
- Numeric example: In triangle ABC, AB = 12 cm and AC = 18 cm. A line through D on AB and E on AC is parallel to BC and AD = 4 cm. Find AE. Solution: AD/AB = AE/AC, so 4/12 = AE/18 => AE = 18*(4/12) = 6 cm.
- Measuring height using shadow (real life): A vertical stick of known height casts a shadow. A taller tree casts a longer shadow. By placing the stick and tree in sunlight so their shadows form similar right triangles, height of the tree / shadow of tree = height of stick / shadow of stick. Rearranging gives the tree's height.
- Scale drawings and maps: A small-scale drawing of a triangular plot can be obtained by drawing a line parallel to the base at a chosen proportion of the sides. All lengths scale proportionally due to Thales' theorem.
- Photography / perspective: When photographing a triangular object, lines parallel to the base produce similar smaller triangles within the frame; proportions in the image reflect actual proportions if the lines are parallel.
- If DE || BC in triangle ABC (D on AB, E on AC): AD/DB = AE/EC
- Equivalent proportional forms: AD/AB = AE/AC = DE/BC
- From AD/AB = AE/AC you can derive AD = AB * (AE/AC) and AE = AC * (AD/AB)
- For n parallel lines cutting AB at A = X0, X1, X2, ... and AC at Y0, Y1, Y2, ... : Xk/Xm = Yk/Ym for corresponding segment sums (proportional division).
Areas of triangles
What is area? Area is the measure of the region enclosed by a figure. For a triangle it is the amount of space inside the three sides, measured in square units (cm², m², etc.).
Base and corresponding height (altitude)
In a triangle any side may be taken as the base. The corresponding height (altitude) is the perpendicular distance from the opposite vertex to the line containing that base. The height must be perpendicular to the chosen base.
Basic formula and reason
If a triangle has base b and corresponding height h, its area is
Area = (1/2) × base × height = 1/2 · b · h
This follows because two congruent copies of the triangle can form a parallelogram whose area is base × height, so each triangle is half of that parallelogram.
Important properties (useful in proofs and problems)
- Triangles on the same base and between the same parallel lines have equal areas.
- For triangles with equal heights, areas are proportional to their bases: Area1 / Area2 = base1 / base2.
- A median (a line from a vertex to the midpoint of the opposite side) divides a triangle into two triangles of equal area.
- If two triangles have equal areas and share the same base, their corresponding heights are equal.
Extensions (useful formulas)
There are other area formulas that are useful in different situations:
- Using two sides and included angle: Area = (1/2) · a · b · sin(C), where a and b are sides and C is the included angle.
- Using Heron's formula (when three sides are known): Let s = (a + b + c)/2 (semiperimeter). Then Area = sqrt[s(s − a)(s − b)(s − c)].
Units
Always express area in square units. If base and height are in metres, area is in m²; if in cm, area is in cm².
- Triangular garden bed: If the base of a triangular flower bed is 4 m and its height is 3 m, area = 1/2 × 4 × 3 = 6 m². This gives how much soil is needed.
- Rooftop gable: A triangular gable of roof with base 8 m and height 2.5 m has area = 1/2 × 8 × 2.5 = 10 m²; used to estimate painting area.
- Comparing two triangular plots with same height: If heights are 10 m and bases are 7 m and 5 m, their areas are proportional to bases (7:5).
- Median property: A triangle with base 12 cm, the median to that base divides it into two triangles each of area = (1/2) × original area.
- Using two sides and included angle: A triangular sail with adjacent sides 5 m and 6 m and included angle 60° has area = 1/2 × 5 × 6 × sin60° ≈ 12.99 m².
- Area = 1/2 × base × height = 1/2 · b · h
- If heights are equal: Area1 / Area2 = base1 / base2
- If bases are equal: Area1 / Area2 = height1 / height2
- Median divides area into two equal parts: Area(Δ) = Area(Δ1) + Area(Δ2) and Area(Δ1) = Area(Δ2)
- Area using two sides and included angle: Area = 1/2 · a · b · sin(C)
- Heron's formula: s = (a + b + c)/2, Area = √[s(s − a)(s − b)(s − c)]
Proof techniques and problem-solving strategies
Overview
In the Class 9 chapter on Triangles, proofs and problem solving rely on careful diagramming, known triangle properties, and standard methods of logical argument. The common proof techniques are: direct proof (use given facts and known theorems), proof by construction (add lines/points), proof by contradiction (assume opposite and find contradiction), and use of congruence and similarity criteria (SSS, SAS, ASA, RHS, AA).
Step-by-step strategy for solving triangle proofs
- Draw a clear diagram: Accurate sketch with labels (vertices A, B, C), tick-marks for equal sides, little arcs for equal angles, and right-angle boxes where needed.
- Mark known facts: Write down what is given and what to prove. Mark equal sides/angles, parallel lines, midpoints, etc.
- Consider auxiliary constructions: Add medians, altitudes, angle bisectors or extend lines if they create congruent or similar triangles.
- Choose a method: Look for congruence (to prove equal angles/sides) or similarity (to relate lengths). Use angle chasing and the triangle-angle-sum property (sum = 180°) or the exterior-angle theorem when needed.
- Write the argument logically: Each step should follow from previous statements or standard theorems. Conclude explicitly what was to be proved.
- Check special cases: e.g., right triangles, isosceles cases, or degenerate positions that need separate handling.
Common small techniques used inside proofs
- Angle chasing: Use angle-sum and exterior angle relations to find unknown angles.
- Tick-marking: Mark equal sides to spot SSS or SAS congruence quickly.
- Use midpoints and medians: Create triangles that share a side and may be congruent.
- Parallel-line method: Draw a line parallel to a side to create corresponding or alternate angles for similarity.
- RHS for right triangles: Use right angle, hypotenuse, and one side equal to conclude congruence.
How to present a proof in answers/exams
- State what is given and what is to be proved.
- Draw and label the figure clearly.
- List constructions you make (for example, "Draw AD ⟂ BC").
- Give reasons for each step (definitions, theorems, congruence or similarity criteria).
- Conclude with a clear statement that the required result is proved.
- Example 1 — Base angles of an isosceles triangle: Given triangle ABC with AB = AC. To prove: ∠ABC = ∠BCA. Construction: Draw altitude AD from A to BC (or median). Reasoning: In triangles ABD and ACD, AB = AC (given), AD = AD (common), and BD = DC if AD is a median. Or use SAS (AB = AC, ∠BAD = ∠CAD because AD is angle bisector). Conclude triangles ABD and ACD are congruent so ∠ABC = ∠BCA. (This uses construction + congruence.)
- Example 2 — Perpendicular from apex of isosceles triangle bisects base and angle: Given AB = AC in triangle ABC, drop AD ⟂ BC. Show D is midpoint of BC and AD bisects ∠A. Triangles ABD and ACD have AB = AC, AD common and right angles at D, use RHS to conclude congruence, so BD = DC and ∠BAD = ∠DAC.
- Example 3 — Height of a tree using similar triangles (real life): A 2 m pole casts a 3 m shadow. At the same time, a tree casts a 15 m shadow. Assuming sun rays are parallel, the pole and tree form similar right triangles. Let tree height = h. Then h/15 = 2/3, so h = (15×2)/3 = 10 m. (This uses similarity and proportionality.)
- Angle-sum property: In triangle ABC, ∠A + ∠B + ∠C = 180°.
- Exterior-angle theorem: An exterior angle equals the sum of the two remote interior angles.
- Triangle inequality (useful heuristic): Any side of a triangle is less than the sum of the other two sides and greater than their difference.
- Congruence criteria: SSS, SAS, ASA, RHS (for right triangles). Use these to prove corresponding sides/angles equal.
- Similarity criteria for triangles: AA (or AAA), SAS (with proportional sides), SSS (with proportional sides). If triangles are similar, corresponding sides are proportional and corresponding angles equal.
- Proportionality result: If a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally (Basic proportionality theorem).
Key Concepts
- Triangle
- A polygon with three sides and three vertices; interior angles sum to 180°.
- Vertex
- A corner point where two sides of a triangle meet.
- Side
- A line segment joining two vertices of a triangle.
- Angle
- The figure formed by two rays or sides with a common endpoint (vertex); measured in degrees.
- Interior angle
- An angle inside a triangle formed by two adjacent sides.
- Exterior angle
- An angle formed by one side of a triangle and the extension of an adjacent side; equals sum of the two opposite interior angles.
- Equilateral triangle
- A triangle with all three sides equal and all three angles 60°.
- Isosceles triangle
- A triangle with at least two equal sides; base angles opposite equal sides are equal.
- Scalene triangle
- A triangle with all sides of different lengths and all angles different.
- Right-angled triangle
- A triangle with one angle equal to 90°.
- Acute triangle
- A triangle in which all three interior angles are less than 90°.
- Obtuse triangle
- A triangle with one interior angle greater than 90°.
- Median
- A line segment joining a vertex to the midpoint of the opposite side.
- Altitude (Height)
- A perpendicular drawn from a vertex to the opposite side (or its extension).
- Perpendicular bisector
- A line that is perpendicular to a side and passes through its midpoint.
- Angle bisector
- A ray or line that divides an angle into two equal angles.
- Centroid
- The point of concurrency of the three medians; it is the triangle's center of mass and divides each median in a 2:1 ratio from the vertex.
- Circumcenter
- The point where the perpendicular bisectors of the sides meet; equidistant from the three vertices and center of the circumscribed circle.
- Orthocenter
- The point of concurrency of the three altitudes of a triangle.
- Incenter
- The point where the three internal angle bisectors meet; equidistant from the sides and center of the inscribed circle.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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In triangle ABC, if AB = AC, then which angles are equal? (a) ∠A and ∠B (b) ∠A and ∠C (c) ∠B and ∠C (d) All three / त्रिभुज ABC में यदि AB = AC, तो कौन-से कोण बराबर हैं? (a) ∠A और ∠B (b) ∠A और ∠C (c) ∠B और ∠C (d) तीनों
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(c) ∠B and ∠C / (c) ∠B और ∠C — समद्विबाहु त्रिभुज में बराबर भुजाओं के सामने के आधार कोण बराबर होते हैं। / In an isosceles triangle, angles opposite the equal sides are equal. Since AB = AC, the base angles ∠B and ∠C are equal.
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Which congruence criterion requires two sides and the included angle? (a) SSS (b) SAS (c) ASA (d) RHS / कौन-सा सर्वांगसमता मानदंड दो भुजाएँ और अंतर्भुज कोण की आवश्यकता रखता है? (a) SSS (b) SAS (c) ASA (d) RHS
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(b) SAS / (b) SAS (Side-Angle-Side) — इसमें दो भुजाएँ और उनके बीच का कोण बराबर होना आवश्यक है। / SAS (Side-Angle-Side) requires two sides and the angle between them to be equal in both triangles.
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Can a triangle have sides of length 2 cm, 3 cm, and 6 cm? (a) Yes (b) No (c) Only if it is right-angled (d) Only if it is obtuse / क्या 2 cm, 3 cm और 6 cm भुजाओं वाला त्रिभुज बन सकता है? (a) हाँ (b) नहीं (c) केवल यदि यह समकोण हो (d) केवल यदि यह अधिककोण हो
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(b) No / (b) नहीं — 2 + 3 = 5 < 6; त्रिभुज असमानता (a + b > c) का उल्लंघन होता है। / The triangle inequality requires the sum of any two sides to exceed the third side: 2 + 3 = 5 < 6, so no triangle is possible.
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In right triangle ABC with the right angle at C, if AC = 3 and BC = 4, what is AB (the hypotenuse)? / समकोण त्रिभुज ABC में C पर समकोण है। AC = 3 और BC = 4 हो, तो AB (कर्ण) क्या होगा?
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AB = 5 / AB = 5 — पाइथागोरस प्रमेय: AB² = AC² + BC² = 9 + 16 = 25 → AB = 5. / By the Pythagorean theorem: AB² = 3² + 4² = 9 + 16 = 25, so AB = 5.
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In triangle ABC, ∠A = 50° and ∠B = 70°. The value of the exterior angle at C is ______°. / त्रिभुज ABC में ∠A = 50° और ∠B = 70° है। C पर बाह्य कोण ______ ° होगा।
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120° / 120° — बाह्य कोण = दोनों असंबद्ध आंतरिक कोणों का योग = ∠A + ∠B = 50° + 70° = 120°. / By the Exterior Angle Theorem, exterior angle at C = ∠A + ∠B = 50° + 70° = 120°.
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True or False: In a triangle, the side opposite the largest angle is the longest side. / सत्य या असत्य: त्रिभुज में सबसे बड़े कोण के सामने की भुजा सबसे लंबी होती है।
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True / सत्य — भुजा-कोण संबंध: यदि ∠A > ∠B, तो a > b. अर्थात् बड़े कोण के सामने लंबी भुजा होती है। / The side-angle relation states: if ∠A > ∠B, then side a > side b. Hence the side opposite the largest angle is the longest.
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State the ASA congruence criterion and give one real-life application of congruent triangles. / ASA सर्वांगसमता मानदंड लिखें और सर्वांगसम त्रिभुजों का एक वास्तविक जीवन उदाहरण दें।
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ASA: If two angles and the included side of one triangle equal the corresponding two angles and included side of another triangle, the triangles are congruent. Real-life: Identical triangular roof trusses in a building use congruence to ensure equal strength and fit. / ASA: यदि एक त्रिभुज के दो कोण और उनके बीच की भुजा दूसरे त्रिभुज के संगत दो कोणों और भुजा के बराबर हों, तो त्रिभुज सर्वांगसम हैं। वास्तविक उदाहरण: एक इमारत में समान त्रिकोणीय ट्रस।
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In triangle ABC, the median is drawn from A to the midpoint M of BC. How does this median divide the area of the triangle? / त्रिभुज ABC में A से BC के मध्य बिंदु M तक माध्यिका खींची जाती है। यह माध्यिका त्रिभुज के क्षेत्रफल को कैसे विभाजित करती है?
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The median divides the triangle into two triangles of equal area. / माध्यिका त्रिभुज को दो बराबर क्षेत्रफल के त्रिभुजों में विभाजित करती है। Since BM = MC (M is the midpoint), both triangles ABM and ACM have equal bases and the same height from A, so area(ABM) = area(ACM).
Related Laws & Principles
Explore allFoundational laws & principles behind this chapter. Each one opens a full page — what it says, why it matters, five practice questions and the mistakes to avoid.