### Types of Triangles:

A triangle can be defined as a polygon which has three angles and three sides. The interior angles of a triangle sum up to 180 degrees and the exterior angles sum up to 360 degrees.

- Scalene Triangle – three sides of a triangle are of different measure
- Isosceles Triangle – two sides are of equal length
- Equilateral Triangle – three sides of a triangle are equal and each of the measures of the interior angles 60 degrees
- Acute angled Triangle – all the angles are smaller than 90 degrees
- Right angle Triangle – has only one 90 degrees angle
- Obtuse-angled Triangle – has an angle which is greater than 90 degrees

**SIMILAR FIGURES**

- Two figures having the same shape but not necessary the same size are called similar figures.
- All congruent figures are similar but all similar figures are not congruent.

**SIMILAR POLYGONS**

Two polygons are said to be similar to each other, if:

(i) their corresponding angles are equal, and

(ii) the lengths of their corresponding sides are proportional

### Similarity Criteria of Triangles

To find whether the given two triangles are similar or not, it has four criteria. They are:

**Side-Side- Side (SSS) Similarity Criterion –**When the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal and the triangle will be considered as similar triangles.**Angle Angle Angle (AAA) Similarity Criterion**– When the corresponding angles of any two triangles are equal, then their corresponding side will be in the same ratio and the triangles are considered to be similar.

**Angle-Angle (AA) Similarity Criterion –**When two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are considered as similar.

**Side-Angle-Side (SAS) Similarity Criterion –**When one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are said to be similar.

**THALES THEOREM OR BASIC PROPORTIONALITY THEORY**

**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.**

Given: In ∆ABC, DE || BC.

To prove:

Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.

Proof: In ∆ADE and ∆BDE,

……..(i) [Area of ∆ = x base x corresponding altitude

In ∆ADE and ∆CDE,

∵ DE || BC …[Given

∴ ar(∆BDE) = ar(∆CDE)

…[∵ As on the same base and between the same parallel sides are equal in area

From (i), (ii) and (iii),

**SIMILAR FIGURES**

- Two figures having the same shape but not necessary the same size are called similar figures.
- All congruent figures are similar but all similar figures are not congruent.

**SIMILAR POLYGONS**

Two polygons are said to be similar to each other, if:

(i) their corresponding angles are equal, and

(ii) the lengths of their corresponding sides are proportional

**Example:**

Any two line segments are similar since length are proportional

Any two circles are similar since radii are proportional

Any two squares are similar since corresponding angles are equal and lengths are proportional.

**Note:**

Similar figures are congruent if there is one to one correspondence between the figures.

∴ From above we deduce:

Any two triangles are similar, if their

(i) Corresponding angles are equal

∠A = ∠P

∠B = ∠Q

∠C = ∠R

(ii) Corresponding sides are proportional

**THALES THEOREM OR BASIC PROPORTIONALITY THEORY**

**Theorem 1:**

**State and prove Thales’ Theorem.**

**Statement:**

**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.**

Given: In ∆ABC, DE || BC.

To prove:

Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.

Proof: In ∆ADE and ∆BDE,

……..(i) [Area of ∆ = x base x corresponding altitude

In ∆ADE and ∆CDE,

∵ DE || BC …[Given

∴ ar(∆BDE) = ar(∆CDE)

…[∵ As on the same base and between the same parallel sides are equal in area

From (i), (ii) and (iii),

**CRITERION FOR SIMILARITY OF TRIANGLES**

**Two triangles are similar if either of the following three criterion’s are satisfied:**

**AAA similarity Criterion.**If two triangles are equiangular, then they are similar.**Corollary(AA similarity).**If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.**SSS Similarity Criterion.**If the corresponding sides of two triangles are proportional, then they are similar.**SAS Similarity Criterion.**If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

**Results in Similar Triangles based on Similarity Criterion:**

- Ratio of corresponding sides = Ratio of corresponding perimeters
- Ratio of corresponding sides = Ratio of corresponding medians
- Ratio of corresponding sides = Ratio of corresponding altitudes
- Ratio of corresponding sides = Ratio of corresponding angle bisector segments.

**AREA OF SIMILAR TRIANGLES**

**Theorem 2.**

**The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding side**

**Results based on Area Theorem:**

- Ratio of areas of two similar triangles = Ratio of squares of corresponding altitudes
- Ratio of areas of two similar triangles = Ratio of squares of corresponding medians
- Ratio of areas of two similar triangles = Ratio of squares of corresponding angle bisector segments.

**Note:**

If the areas of two similar triangles are equal, the triangles are congruent.

**PYTHAGORAS**** THEOREM:**

**In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.**

Given: ∆ABC is a right triangle right-angled at B.

To prove: AB² + BC² = AC²

Const.: Draw BD ⊥ AC

Proof: In ∆s ABC and ADB,

∠A = ∠A …[common

∠ABC = ∠ADB …[each 90°

∴ ∆ABC ~ ∆ADB …[AA Similarity

∴ ………[sides are proportional]

⇒ AB² = AC.AD

Now in ∆ABC and ∆BDC

∠C = ∠C …..[common]

∠ABC = ∠BDC ….[each 90°]

∴ ∆ABC ~ ∆BDC …..[AA similarity]

∴ ……..[sides are proportional]

BC² = AC.DC …(ii)

On adding (i) and (ii), we get

AB² + BC² = ACAD + AC.DC

⇒ AB² + BC² = AC.(AD + DC)

AB² + BC² = AC.AC

∴AB² + BC² = AC²