Chapter 6 — Triangles

Triangle — definition: A triangle is a polygon with three sides and three vertices. If the vertices are A, B and C, the sides are AB, BC and CA and the interior angles are <A, <B and <C.

Basic elements:

• Vertex: a corner point (A, B, C).
• Side: the line segment between two vertices (e.g. AB).
• Interior angle: angle formed at a vertex inside the triangle (e.g. <A).
• Exterior angle: angle formed by extending a side (an exterior angle at A is supplementary to <A).

Types of triangles by sides: equilateral (all sides equal), isosceles (two sides equal), scalene (all sides different).

Types by angles: acute (all angles < 90°), right (one angle = 90°), obtuse (one angle > 90°).

Special line segments:

• Median: segment joining a vertex to the midpoint of the opposite side.
• Altitude (height): perpendicular from a vertex to the opposite side (or its extension).
• Angle bisector: line that divides an angle into two equal angles.
• Perpendicular bisector: line perpendicular to a side at its midpoint.

Important points (centers):

• Centroid (G): intersection of medians. It is the triangle's centre of mass; each median is divided by G in a 2:1 ratio (vertex to centroid : centroid to midpoint = 2:1).
• Circumcenter (O): intersection of perpendicular bisectors. Center of the circumcircle; equidistant from the vertices. It may lie inside, on, or outside the triangle (acute, right, obtuse respectively).
• Incenter (I): intersection of angle bisectors. Center of the incircle; equidistant from the sides and always inside the triangle.
• Orthocenter (H): intersection of altitudes. Can lie inside, on, or outside the triangle.

Basic properties and theorems:

• Sum of interior angles: <A + <B + <C = 180°.
• Exterior angle theorem: An exterior angle equals the sum of the two opposite interior angles (e.g., exterior at A = <B + <C).
• Triangle inequality: Sum of lengths of any two sides is greater than the third: a + b > c, b + c > a, c + a > b.
• Opposite relations: Larger side is opposite larger angle and vice versa.

Area and perimeter (basic): Perimeter P = a + b + c. Area can be found by (1/2) × base × height. For a right triangle with legs a and b and hypotenuse c, area = (1/2)ab and a^2 + b^2 = c^2.

This collection of definitions and properties forms the foundation for solving construction, congruence, similarity and mensuration problems involving triangles.

Definition: Two figures are similar if one can be obtained from the other by a sequence of rigid motions (translation, rotation, reflection) and uniform scaling. For triangles, similarity means all corresponding angles are equal and the lengths of corresponding sides are proportional.

Key ideas for triangles:

• Corresponding angles equal: If ∆ABC ~ ∆DEF then ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
• Corresponding sides proportional: AB/DE = BC/EF = CA/FD = k, where k is the scale factor (ratio of similarity).
• Consequences: Ratios of perimeters = k, ratios of corresponding altitudes/medians/angle-bisectors = k, and ratio of areas = k^2.

Criteria (tests) for similarity of triangles:

• AA (Angle-Angle): If two angles of one triangle are equal to two angles of another, the triangles are similar.
• SAS (Side-Angle-Side) for 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): If all three pairs of corresponding sides are in proportion, the triangles are similar.

Using similarity: Similarity is widely used to solve problems involving indirect measurement (heights and distances), map scaling, architectural models, lens/camera scaling, and proving geometrical results.

Remarks on similar polygons and 3D shapes: For similar polygons of n sides, corresponding angles are equal and corresponding side lengths are proportional. For similar solids (3D), linear dimensions scale by k, areas scale by k^2 and volumes scale by k^3.

What is similarity?
Two figures are similar if they have the same shape though not necessarily the same size. For triangles, similarity means their corresponding angles are equal and corresponding sides are in the same ratio (proportional).

Notation: ΔABC ~ ΔDEF means triangle ABC is similar to triangle DEF. Then ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD.

Three main criteria (CBSE Class 10):

• AA (Angle-Angle): If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. (Two equal angles force the third to 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, then the triangles are similar. Formally, if AB/DE = AC/DF and ∠A = ∠D, then ΔABC ~ ΔDEF.
• SSS (Side-Side-Side) similarity: If the three sides of one triangle are proportional to the three sides of another triangle, then the triangles are similar. That is, if AB/DE = BC/EF = CA/FD, then ΔABC ~ ΔDEF.

Why these work – short proofs/outlines:

• AA: If two angles match, the third must match (sum of angles = 180°). Equal angles determine equal shape → similarity.
• SAS: If two sides are in proportion and the included angle is equal, you can scale one triangle about the included angle so the two pairs of sides coincide in length; corresponding third sides then also match proportionally, giving similarity.
• SSS: If all three side ratios are equal, construct a scaled copy of one triangle with that ratio; corresponding angles coincide, giving similarity.

How to use these criteria in problems (step-by-step):

1. Identify or mark equal angles and proportional sides.
2. Match corresponding vertices (this determines corresponding sides and angles).
3. Apply the appropriate criterion (AA, SAS, SSS).
4. Find the scale factor k (ratio of corresponding sides). Use it to get other lengths: perimeter scales by k, area scales by k².

Consequences and useful results:

• If ΔABC ~ ΔDEF with scale factor k = AB/DE, then corresponding heights, medians and angle bisectors are in ratio k.
• Perimeter(ABC)/Perimeter(DEF) = k.
• Area(ABC)/Area(DEF) = k².
• If a line drawn parallel to one side of a triangle cuts the other two sides, the smaller triangle formed is similar to the original (Basic Proportionality Theorem / Thales).

Tips for identifying corresponding parts: Corresponding angles are matched first (mark them). The sides opposite equal angles are corresponding sides. Always keep vertex order consistent when writing similarity (e.g., ΔABC ~ ΔPQR means A↔P, B↔Q, C↔R).

Statement (Basic Proportionality Theorem / Thales): If a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those sides in the same ratio.

More precisely, in triangle ABC if a line through D on AB and E on AC is parallel to BC (DE ∥ BC), then

AD / DB = AE / EC.

Proof (using similar triangles):

1. Given DE ∥ BC, angles ∠ADE = ∠ABC and ∠AED = ∠ACB (corresponding angles).
2. So, ΔADE is similar to ΔABC (AA similarity).
3. From similarity, AD/AB = AE/AC = DE/BC. From AD/AB = AE/AC we get AD/(AB−AD) = AE/(AC−AE), i.e. AD/DB = AE/EC.

Converse: If a line through points D on AB and E on AC (with D and E between the vertices) divides AB and AC proportionally, i.e., AD/DB = AE/EC, then DE ∥ BC.

Proof of converse (construction / contradiction): Through D draw a line D E' parallel to BC meeting AC at E'. By the Basic Proportionality Theorem AD/DB = AE'/E'C. If AD/DB = AE/EC as given, then AE' = AE and hence E' = E. Thus DE ∥ BC.

Notes and special cases:

• Points D and E must lie on the segments AB and AC (or appropriate extensions—extended versions consider rays).
• Mid-segment corollary: If D and E are mid-points of AB and AC respectively, then DE ∥ BC and DE = (1/2)·BC.

Why this matters: The theorem is a basic tool to create and identify similar triangles, solve proportional length problems, and is widely used in surveying, map scaling and design.

What it means
If two triangles are similar, their corresponding angles are equal and corresponding sides are in the same ratio (called the similarity ratio). From this basic fact follow several important consequences used frequently in Class 10 geometry.

Main consequences (with brief justification)

• Corresponding sides are proportional: If ΔABC ~ ΔA'B'C' then AB/A'B' = BC/B'C' = CA/C'A'. (Direct from similarity.)
• Perimeter ratio equals side ratio: Perimeter(ΔABC)/Perimeter(ΔA'B'C') = AB/A'B' (sum of corresponding sides).
• Area ratio equals square of similarity ratio: Area(ΔABC)/Area(ΔA'B'C') = (AB/A'B')^2. (Because area ∝ product of two side lengths and both scale by the similarity ratio.)
• Basic Proportionality Theorem (Thales): If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally. (If DE ∥ BC in ΔABC with D on AB and E on AC, then AD/DB = AE/EC.)
• Angle bisector theorem: The internal bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides. (If AD bisects ∠A in ΔABC and D is on BC, then BD/DC = AB/AC.)
• Right-triangle altitude to hypotenuse (consequences of similarity): In right ΔABC with ∠C = 90° and altitude from C meeting hypotenuse AB at D,
  • ΔCAD ~ ΔCAB and ΔCBD ~ ΔCBA, so AC^2 = AB·AD and BC^2 = AB·BD.
  • Also CD^2 = AD·BD and CD = (AC·BC)/AB.
  (These follow because the three right-related triangles are all similar to each other.)

How to use these consequences
Most problems reduce to identifying pairs of similar triangles (often by AA), writing ratios of corresponding sides, and then using those ratios to find unknown lengths or areas. For right-triangle altitude problems use the special altitude relations above; for angle bisector problems use the bisector ratio; when a line is parallel use Thales' proportionality.

Overview
When a perpendicular is drawn from the right angle of a right triangle to its hypotenuse, the altitude divides the original triangle into two smaller right triangles. Each of these small triangles is similar to the original triangle and to each other. This yields powerful relations between sides, projections on the hypotenuse, and the altitude.

Diagram and notation
Consider right triangle ABC with right angle at C. Let AB be the hypotenuse, AC and BC the two legs. Draw perpendicular CD from C to AB, meeting AB at D. Set AB = c, BC = a, AC = b, AD = p, DB = q, CD = h. Note that p + q = c.

Similarity and proof (sketch)
Triangles ABC, ACD and CBD are all right triangles. Angle A of triangle ABC equals angle A of triangle ACD; angle B of ABC equals angle B of CBD. Thus by AA, the three triangles are similar: ABC ~ ACD ~ CBD. From these similarities we get proportionalities that lead to the altitude-to-hypotenuse results below.

Key derived relations
From similarity ABC ~ ACD: b / c = p / b, so b² = c·p.
From similarity ABC ~ CBD: a / c = q / a, so a² = c·q.
From similarity ACD ~ CBD: h / p = q / h, so h² = p·q.

Combining these gives further useful formulas: h = (a·b) / c, and p = b² / c, q = a² / c. These relate legs, hypotenuse and projections. The Pythagorean relation a² + b² = c² is consistent with p + q = c.

Geometric-mean interpretations
Each leg is the geometric mean of the hypotenuse and the adjacent projection: AC = sqrt(AB·AD), BC = sqrt(AB·DB). The altitude CD is the geometric mean of the two hypotenuse segments: CD = sqrt(AD·DB).

Why useful
These results let you compute unknown lengths in right triangles when an altitude or projection is known, solve construction problems, and appear in many real-life measurements and designs where right triangles occur.

Definition and main result

Two triangles are called similar if their corresponding angles are equal and corresponding sides are in proportion. If two triangles ΔABC and ΔDEF are similar with corresponding sides AB/DE = BC/EF = CA/FD = k (the scale factor), then their areas are in the ratio of the squares of corresponding sides.

Theorem: If ΔABC ~ ΔDEF and the ratio of any pair of corresponding sides is k, then
Area(ΔABC) / Area(ΔDEF) = k².

Proof (using corresponding base and height)

1. Let AB correspond to DE. Put base AB of ΔABC and base DE of ΔDEF. Because triangles are similar, the perpendicular heights to these bases (from the opposite vertices) are in the same ratio k as the bases.
2. Area(ΔABC) = (1/2)·AB·(height to AB). Area(ΔDEF) = (1/2)·DE·(height to DE).
3. Divide the two areas: Area(ΔABC)/Area(ΔDEF) = [AB/DE]·[height to AB / height to DE] = k·k = k².

Corollaries and remarks

• Area ratios equal the square of the ratio of any corresponding linear measure (sides, medians, altitudes, inradius, circumradius). Example: (area1/area2) = (median1/median2)².
• Perimeter ratio = side ratio = k, so area ratio = (perimeter ratio)².
• More generally, for any two similar plane figures, areas scale as the square of the linear scale factor.

Short example of application

If a model of a triangular park is made at scale 1:200 and the actual park has area A, then the model area is A/(200²) because area scales by the square of the linear scale factor.

In Class 10 Triangles, the main ideas (similarity, Basic Proportionality Theorem, Pythagoras theorem and area relations) are used in many practical problems. Typical applications include finding unknown lengths or heights, proving equal ratios, comparing areas of triangles, and solving geometry problems by reducing them to similar triangles or right-triangle problems.

Key strategies used repeatedly:

• Recognise similar triangles (usually by AA — equal angles) and set up side ratios.
• Apply the Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides proportionally.
• Use the Pythagorean theorem in right triangles to find a missing side.
• Use area relations: if two triangles are on the same base (or between same parallels), compare areas via base or height; if triangles are similar, their areas are in the square of the similarity ratio.
• Introduce an altitude or draw auxiliary lines to create right triangles or pairs of similar triangles.

Typical problem types (brief):

• Height and distance problems: measure heights of towers/trees using similar triangles formed by shadows or sight lines.
• Finding segments: use the Basic Proportionality Theorem to find unknown segments when a line is parallel to a side of a triangle.
• Determining unknown sides using similarity: identify two similar triangles and compute missing lengths from proportionality.
• Area-comparison problems: relate areas of triangles that are similar or have same base/height to find unknown measures.
• Pythagorean problems: direct application in right triangles, or use to check right angles and compute diagonals.

Exam approach: Draw a clear diagram, mark equal angles or parallelism, name similar triangles, write side ratios, solve algebraically. Where appropriate, state which theorem or test (AA, BPT, Pythagoras) you used.

Overview
In the chapter "Triangles" theorems are often supported by short results called lemmas, corollaries and auxiliary results. These are small but powerful statements used repeatedly to prove bigger results (like similarity, congruence, area relations). Below are the most important statements, short explanations and why they matter.

1. Basic Proportionality Theorem (BPT) / Thales' theorem
If a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally. If DE ∥ BC in ΔABC with D on AB and E on AC, then AD/DB = AE/EC. This is used to create similar triangles and to relate segment ratios.

Converse of BPT
If a line cuts two sides of a triangle and divides them proportionally, then the line is parallel to the third side.

2. Midpoint theorem
If M and N are midpoints of AB and AC in ΔABC, then MN ∥ BC and MN = (1/2) BC. Corollaries: A segment joining midpoints in a triangle is parallel to the third side and half its length; repeated application gives smaller similar triangles.

3. Similarity criteria and related lemmas
- AA (two equal angles) ⇒ triangles similar.
- SAS (two sides in proportion and included angle equal) ⇒ similar.
- SSS (three sides in proportion) ⇒ similar.
Auxiliary lemma: corresponding medians, altitudes and angle bisectors of similar triangles are in the same ratio as corresponding sides.

4. Area relations / same-base-same-parallels lemma
If two triangles have the same base and lie between the same parallels, they have equal areas. More generally, for triangles with equal altitudes, areas are proportional to their bases. For similar triangles, area ratio = (ratio of corresponding sides)^2.

5. Exterior angle theorem and angle-side results
An exterior angle of a triangle equals the sum of the two opposite interior angles. Also, the greater side of a triangle is opposite the greater angle and conversely.

6. Right-triangle auxiliary results
If an altitude is drawn from the right angle to the hypotenuse in a right triangle, the two smaller triangles are similar to the original triangle and to each other. Also, the altitude^2 = product of the projections on the hypotenuse; each leg^2 = (hypotenuse) × (projection of that leg on hypotenuse). Another useful fact: the median to the hypotenuse equals half the hypotenuse and is the circumradius of the right triangle.

How these are used
These lemmas are tools: use BPT or midpoint theorem to introduce auxiliary parallel lines to create similar triangles; use area lemmas to compare areas; use right-triangle similarity results to find lengths after dropping altitudes. Problems in the chapter typically require adding one auxiliary line (parallel, median, altitude) and then applying one or more of these lemmas to proceed.

What this topic is about
"Skills and proof techniques" in Class 10 Triangles trains you to convert geometric facts into clear logical proofs. You learn to identify what is given, decide which theorems or congruence/similarity criteria apply, draw helpful auxiliary lines, and present a short, correct sequence of statements and reasons.

Steps for writing a clear triangle proof

• Read carefully: Note exactly what is given and what must be proved.
• Plan: Ask which known result helps (congruence, similarity, angle sum, exterior angle, etc.).
• Construct if needed: Draw an auxiliary line (median, altitude, perpendicular bisector, or extension) to create congruent/similar triangles or right triangles.
• Use a criterion: Apply SSS/SAS/ASA/AAS/RHS (or similarity criteria) to deduce equality of angles/sides.
• Conclude: State clearly how the intermediate results give the required result.
• Check: Ensure each statement has a justified reason (definition, theorem, or congruence).

Common techniques (with short descriptions)

• Auxiliary line: Draw a median, altitude, or extend a side to form useful triangles.
• Congruence approach: Show two triangles are congruent (SSS, SAS, ASA, AAS, RHS) and transfer equal parts.
• Angle-chasing: Express unknown angles using given ones and known relations (linear pair, vertically opposite, exterior angle = sum of two remote interior angles).
• Use of symmetry / isosceles facts: In an isosceles triangle, equal sides imply equal base angles; perpendicular from apex to base bisects base.
• Contradiction: Assume the opposite of what you want and derive an impossibility (use sparingly at this level).
• Right-triangle tools: Use RHS or Pythagoras to link side lengths in right triangles.

How to present a short proof

1. Begin with "Given" (list givens).
2. Write "To prove" (the statement).
3. Give a short construction if you add a line.
4. Provide 3–6 numbered statements with brief reasons (e.g., "ΔABC ≅ ΔDEF (SAS)"), then the final conclusion.

Tips to avoid common mistakes

• Don’t assume equality without reason; always cite a theorem or congruence.
• Choose the congruence criterion that uses the most readily available data.
• Label equal angles/sides on your diagram to avoid confusion during angle-chasing.