## CBSE Class 10 Maths Notes Chapter 6 Triangles

**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

\(\frac { AB }{ PQ } =\frac { AC }{ PR } =\frac { BC }{ QR } \)

**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: \(\frac { AD }{ DB } =\frac { AE }{ EC } \)

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

Proof: In ∆ADE and ∆BDE,

\(\frac { ar(\Delta ADE) }{ ar(\Delta BDE) } =\frac { \frac { 1 }{ 2 } \times AD\times EM }{ \frac { 1 }{ 2 } \times DB\times EM } =\frac { AD }{ DB } \) ……..(i) [Area of ∆ = \(\frac { 1 }{ 2 }\) x base x corresponding altitude

In ∆ADE and ∆CDE,

\(\frac { ar(\Delta ADE) }{ ar(\Delta CDE) } =\frac { \frac { 1 }{ 2 } \times AE\times DN }{ \frac { 1 }{ 2 } \times EC\times DN } =\frac { AE }{ EC } \)

∵ 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),

\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

**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 sides.**

Given: ∆ABC ~ ∆DEF

To prove: \(\frac { ar(\Delta ABC) }{ ar(\Delta DEF) } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } \)

Const.: Draw AM ⊥ BC and DN ⊥ EF.

Proof: In ∆ABC and ∆DEF

\(\frac { ar(\Delta ABC) }{ ar(\Delta DEF) } =\frac { \frac { 1 }{ 2 } \times BC\times AM }{ \frac { 1 }{ 2 } \times EF\times DN } =\frac { BC }{ EF } .\frac { AM }{ DN } \) …(i) ……[Area of ∆ = \(\frac { 1 }{ 2 }\) x base x corresponding altitude

∵ ∆ABC ~ ∆DEF

∴ \(\frac { AB }{ DE } =\frac { BC }{ EF } \) …..(ii) …[Sides are proportional

∠B = ∠E ……..[∵ ∆ABC ~ ∆DEF

∠M = ∠N …..[each 90°

∴ ∆ABM ~ ∆DEN …………[AA similarity

∴ \(\frac { AB }{ DE } =\frac { AM }{ DN } \) …..(iii) …[Sides are proportional

From (ii) and (iii), we have: \(\frac { BC }{ EF } =\frac { AM }{ DN } \) …(iv)

From (i) and (iv), we have: \(\frac { ar(\Delta ABC) }{ ar(\Delta DEF) } =\frac { BC }{ EF } .\frac { BC }{ EF } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \)

Similarly, we can prove that

\(\frac { ar(\Delta ABC) }{ ar(\Delta DEF) } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { AC^{ 2 } }{ DF^{ 2 } } \)

∴\(\frac { ar(\Delta ABC) }{ ar(\Delta DEF) } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } =\frac { AC^{ 2 } }{ DF^{ 2 } } \)

**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**

**Theorem 3:**

**State and prove Pythagoras’ Theorem.**

**Statement:**

**Prove that, 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

∴ \(\frac { AB }{ AD } =\frac { AC }{ AB } \) ………[sides are proportional]
⇒ AB² = AC.AD

Now in ∆ABC and ∆BDC

∠C = ∠C …..[common]
∠ABC = ∠BDC ….[each 90°]
∴ ∆ABC ~ ∆BDC …..[AA similarity]
∴ \(\frac { BC }{ DC } =\frac { AC }{ BC } \) ……..[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

**CONVERSE OF PYTHAGORAS THEOREM**

**Theorem 4:**

**State and prove the converse of Pythagoras’ Theorem.**

**Statement:**

**Prove that, in a triangle, if 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.**

Given: In ∆ABC, AB² + BC² = AC²

To prove: ∠ABC = 90°

Const.: Draw a right angled ∆DEF in which DE = AB and EF = BC

Proof: In ∆ABC,

AB² + BC² = AC² …(i) [given]
In rt. ∆DEF

DE² + EF² = DF² …[by pythagoras theorem]
AB² + BC² = DF² …..(ii) …[DE = AB, EF = BC]
From (i) and (ii), we get

AC² = DF²

⇒ AC = DF

Now, DE = AB …[by cont]
EF = BC …[by cont]
DF = AC …….[proved above]
∴ ∆DEF ≅ ∆ABC ……[sss congruence]
∴ ∠DEF = ∠ABC …..[CPCT]
∠DEF = 90° …[by cont]
∴ ∠ABC = 90°

**Results based on Pythagoras’ Theorem:**

**(i) Result on obtuse Triangles.**

If ∆ABC is an obtuse angled triangle, obtuse angled at B,

If AD ⊥ CB, then

AC² = AB² + BC² + 2 BC.BD

**(ii) Result on Acute Triangles.**

If ∆ABC is an acute angled triangle, acute angled at B, and AD ⊥ BC, then

AC² = AB² + BC² – 2 BD.BC.