Ch- 6 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.

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:

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.

Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.
Proof: In ∆ADE and ∆BDE,

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

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

∵ 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),
$frac { AD }{ DB } =frac { AE }{ EC }$

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,
$frac { ar(Delta ADE) }{ ar(Delta BDE) } =frac { frac { 1 }{ 2 } times ADtimes EM }{ frac { 1 }{ 2 } times DBtimes EM } =frac { AD }{ DB }$ ……..(i) [Area of ∆ = $frac { 1 }{ 2 }$ x base x corresponding altitude
In ∆ADE and ∆CDE,
$frac { ar(Delta ADE) }{ ar(Delta CDE) } =frac { frac { 1 }{ 2 } times AEtimes DN }{ frac { 1 }{ 2 } times ECtimes DN } =frac { AE }{ EC }$
∵ 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 side

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

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.

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

CBSE Class 10th Maths

Types of Triangles:

Similarity Criteria of Triangles

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,

Course Curriculum

CBSE Class 10th Maths

Ch- 6 Triangles

Types of Triangles:

Similarity Criteria of Triangles

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,

Course Curriculum

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,