Triangles

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06 TRIANGLES

Similarity:

Two figures are similar if their shapes are same but size may not be the same.

e.g Tajmahal and model of Tajmahal

Similarity of triangles:

Two triangles Δ ABC and Δ PQR are similar

If 1) Their corresponding angles are equal i.e

∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R

2) Their corresponding sides are in proportion i.e.

\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}

A n d w e d e n o t e i t b y △ A B C ~ △ P Q R

Theorem 1:Basic proportionality Theorem:(Thale’s Theorem)

If a line is drawn parallel to one side of triangle intersecting the other two sides, then

it divides the two sides in the same ratio.

Given: A triangle ABC in which

, and DE intersects AB at D and AC at E.

T O P R O V E : \frac{A D}{D B} = \frac{A E}{A C}

C O N S T R U C T I O N : J o i n B E, C D a n d d r a w E F ⊥ B A a n d D G ⊥ C A .

P r o o f : A r e a (△ A B E) = \frac{1}{2} (b a s e \times h e i g h t) = \frac{1}{2} (A D \times E F) : A r e a (△ D B E) = \frac{1}{2} (b a s e \times h e i g h t) = \frac{1}{2} (D B \times E F)

∴ \frac{A r e a (△ A D E)}{A r e a (△ D B E)} = \frac{\frac{1}{2} (A D \times E F)}{\frac{1}{2} (D B \times E F)} = \frac{A D}{D B} . . . . . . . (1) s i m i l a r l y, \frac{A r e a (△ A D E)}{A r e a (△ D E C)} = \frac{\frac{1}{2} (A E \times D G)}{\frac{1}{2} (E C \times D G)} = \frac{A E}{E C} . . . . . . . (2)

∴ A r e a (△ D B E) = A r e a (△ D E C) [t r i a n g l e s w i t h t h e s a m e b a s e D E a n d b e t w e e n t h e s a m e p a r a l l e l s D E a n d B C .]

\Rightarrow \frac{1}{A r e a (△ D B E)} = \frac{1}{A r e a (△ D E C)} [T a k i n g r e c i p r o c a l s]

\Rightarrow \frac{A r e a (△ A D E)}{A r e a (△ D B E)} = \frac{A r e a (△ A D E)}{A r e a (△ D E C)} [M u l t i p l y i n g b y A r e a]

\Rightarrow \frac{A D}{D B} = \frac{A E}{E C} [U \sin g (1) a n d (2)]

C O R O L L A R Y : I f i n a △ A B C, a l i n e D E B C, i n t e r s e c t s A B a t D a n d A C a t E, t h e n : (i) \frac{A B}{A D} = \frac{A C}{A E} (i i) \frac{A B}{D B} = \frac{A C}{E C}

P R O O F (i) \frac{A D}{D B} = \frac{A E}{E C} [F r o m t h e b a s i c p r o p o r t i o n a l i t y t h e o r e m]

\Rightarrow \frac{D B}{A D} = \frac{E C}{A E} [T a k i n g r e c i p r o c a l s]

\Rightarrow 1 + \frac{D B}{A D} = 1 + \frac{E C}{A E} [A d d i n g 1 o n b o t h s i d e s]

\frac{A D + D B}{A D} = \frac{A E + E C}{A E}

\Rightarrow \frac{A B}{A D} = \frac{A C}{A E}

(i i) \frac{A D}{D B} = \frac{A E}{E C} [F r o m t h e b a s i c p r o p o r t i o n a l i t y t h e o r e m]

\Rightarrow \frac{A D}{D B} = \frac{A E}{E C} [A d d i n g 1 o n b o t h s i d e s]

\Rightarrow \frac{A D + D B}{D B} = \frac{A E + E C}{E C}

\Rightarrow \frac{A B}{D B} = \frac{A E}{E C}

THEOREM 2:(Converse of Basic Proportionality Theorem)

If a line divides any two sides of a triangle in the same ratio, then the line must be

Parallel to the third side.

G I V E N : I n △ A B C, a l i n e D E i n t e r s e c t i n g A B a n d A C a t D, E, r e s p e c t i v e l y .

\frac{A D}{D B} = \frac{A E}{E C}

T O P R O V E : D E ∥ B C

P R O O F : I f p o s s i b l e, l e t D E b e n o t p a r a l l e l t o B C . T h e n, t h e r e m u s t b e a n o t h e r l i n e p a r a l l e l t o B C .

L e t D F ∥ B C

\frac{A D}{D B} = \frac{A F}{F C} (B Y B P T) . . . . . (i)

B u t \frac{A D}{D B} = \frac{A E}{E C} (G I i v e n) . . . . . . . \dots (i i)

From (i) and (ii), we get

\frac{A F}{F C} = \frac{A E}{E C}

\Rightarrow \frac{A F}{A C} + 1 = \frac{A E}{E C} + 1 [A d d i n g 1 o n b o t h s i d e s]

\Rightarrow \frac{A C}{F C} = \frac{A C}{E C}

\Rightarrow E C = E C

T h i s i s p o s s i b l e o n l y w h e n F a n d E c o i n c i d e B u t D F ∥ B C . H e n c e, D F ∥ B C .

THEOREM 3(Angle bisector theorem):

The internal bisector of an angle of a triangle divides the opposite side internally in

the ratio of the sides containing the angle.

G I V E N : I n, A D i s b i s e c t o r o f ∠ A .

T O P R O V E : \frac{B D}{D C} = \frac{A B}{A C}

C O N S T R U C T I O N : D r a w C E ∥ D A . L e t C E m e e t B A p r o d u c e d i n E .

P R O O F : S i n c e C E ∥ D A a n d A C i s t r a n s v e r s a l .

∴ ∠ 2 = ∠ 3 \dots (i) [A l t e r n a t e a n g l e s]

a n d, ∠ 1 = ∠ 4 \dots (i i) [c o r r e s p o n d i n g a n g l e s]

B u t, ∠ 1 = ∠ 2 [∵ A D i s t h e b i s e c t o r o f ∠ A]

From (i) and (ii), we get

∠ 3 = ∠ 4

T h u s, i n △ A C E

\Rightarrow A E = A C \dots (i i i) [S i d e s o p p o s i t e t o e q u a l a n g l e s a r e e q u a l]

N o w, i n △ B C E, D A ∥ C E

\Rightarrow \frac{B D}{D C} = \frac{B A}{A E} [U \sin g B a s i c P r o p o r t i o n a l i t y T h e o r e m]

\Rightarrow \frac{B D}{D C} = \frac{A B}{A C} [A E = A C]

H e n c e, \frac{B D}{D C} = \frac{A B}{A C}

THEOREM 4(Converse of angle bisector theorem):

I n T r i a n g l e A B C, i f D i s a p o i n t o n B C s u c h t h a t, \frac{B D}{D C} = \frac{A B}{A C}

p r o v e t h a t A D i s t h e b i s e c t o r o f ∠ A .

G I V E N : I n A B C, i n w h i c h D i s a p o i n t o n B C s u c h t h a t \frac{B D}{D C} = \frac{A B}{A C}

T O P R O V E : A D i s t h e b i s e c t o r o f ∠ A .

CONSTRUCTION :Produce BA to E such that AE = AC. Join EC

P R O O F : I n △ A B C, w e h a v e A E = A C

\Rightarrow ∠ 3 = ∠ 4

Now,

\frac{B D}{D C} = \frac{A B}{A C}

T h u s, i n △ B C E, w e h a v e

\frac{B D}{D C} = \frac{A B}{A E} [∵ A C = A E]

D A ∥ C E (B y c o n v e r s e o f B P T)

\Rightarrow ∠ 1 = ∠ 4 \dots (i i) [C o r r e s p o n d i n g a n g l e s]

a n d, ∠ 2 = ∠ 3 \dots (i i i) [A l t e r n a t e a n g l e s]

B u t, ∠ 3 = ∠ 4 [F r o m (i)]

∴ ∠ 1 = ∠ 2 . . . . . . [F r o m (i i) a n d (i i i)]

H e n c e, A D i s t h e b i s e c t o r o f ∠ A .

THEOREM 5:

If three or more parallel lines are intersected by two transversals, prove that the

intercepts made by them on the transversals are proportional.

GIVEN:Three parallel lines l, m, n which are cut by the transversals AB and CD at P, Q, R and E, F, G respectively.

T O P R O V E : \frac{P Q}{Q R} = \frac{E F}{F G}

CONSTRUCTION: join PG.

PROOF:

I N △ P R G, Q M ∥ R G \Rightarrow \frac{P Q}{Q R} = \frac{P M}{M G} \dots (i) (B y T h a l e ’ s t h e o r e m)

N o w, i n △ P E G, w e h a v e

F M ∥ P E

\Rightarrow \frac{E F}{F G} = \frac{P M}{M G} . . (i i) (B y B P T)

H e n c e, \frac{P Q}{Q R} = \frac{E F}{F G} . . . . [U \sin g (i) a n d (i i)]

Criteria for similarities of triangles:

THEOREM6:(AAA Similarity Criterion):

If two triangles are equiangular, then they are similar.

G I V E N : T w o t r i a n g l e s A B C a n d D E F s u c h t h a t ∠ A = ∠ D, ∠ B = ∠ E a n d ∠ C = ∠ F

T O P R O V E : △ A B C ∥ △ D E F i . e \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F}

PROOF: We can have AB=DE,,AB>DE,AB<DE

Case I: When AB = DE.

In this case, we have

∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F a n d A B = D E

△ A B C ≅ △ D E F . . . (B y A S A c r i t e r i o n o f c o n g r u e n c y)

\Rightarrow A B = D E, B C = E F a n d A C = D F

\frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F}

Hence proved

C a s e I I : W h e n A B < D E .

Take a point P on the line DE and Q on the line DF such that AB = DP and AC = DQ.

Join PQ.

I n △ A B C a n d △ D P Q,

A B = D P, ∠ A = ∠ D, A C = D Q

∴ △ △ A B C ≅ △ D P Q [B y S A S c r i t e r i o n o f c o n g r u e n c e]

∠ B = ∠ D P Q

b u t ∠ B = ∠ E = ∠ D E F

∴ ∠ D P Q = ∠ D E F

\Rightarrow P Q ∥ E F [C o r r e s p o n d i n g a n g l e s a r e e q u a l]

\frac{D P}{D E} = \frac{D Q}{D F} [B y c o r o l l a r y o f T h a l e ’ s T h e o r e m]

Similarly, we can prove that

\frac{A B}{D E} = \frac{B C}{E F}

\frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F}

Hence proved

Case III:When AB > DE

Take a point P on the line DE produced and Q on the line DF produced

so that DP = AB and DQ = AC.

Join PQ.

I n △ A B C a n d △ D P Q, w e h a v e

A B = D P, A C = D Q, ∠ A = ∠ D

∴ A B C ≅ D P Q [B y S A S c r i t e r i o n o f c o n g r u e n c e]

∠ D = ∠ D P Q

b u t ∠ D = ∠ E = ∠ D E F

\Rightarrow P Q ∥ E F [C o r r e s p o n d i n g a n g l e s a r e e q u a l]

\frac{D E}{D P} = \frac{D F}{D Q} [B y c o r o l l a r y o f T h a l e ’ s T h e o r e m]

\frac{A B}{D E} = \frac{A C}{D F}

Similarly, we can prove that

\frac{A B}{D E} = \frac{B C}{E F}

\frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F}

hence proved

THEOREM 7: (SSS Similarity Criterion)

If the corresponding sides of two triangles are proportional, then they are similar.

G I V E N : T w o t r i a n g l e s A B C a n d D E F s u c h t h a t \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F}

T O P R O V E : △ A B C ∥ △ D E F

CONSTRUCTION:Let P and Q be points on DE and DF respectively

such that DP = AB and DQ = AC.

Join PQ.

PROOF :We have,

\frac{A B}{D E} = \frac{A C}{E F}

\frac{D P}{D E} = \frac{D Q}{D F} [A B = D P a n d A C = D Q]

\Rightarrow P Q ∥ E F [B y t h e c o n v e r s e o f T h a l e ’ s T h e o r e m]

∠ D P Q = ∠ E a n d ∠ D P Q = ∠ F [C o r r e s p o n d i n g a n g l e s]

Thus, in triangles DPQ and DEF, we have

∠ D P Q = ∠ E a n d ∠ D P Q = ∠ F

△ D P Q ∥ ∆ D E F (A - A c r i t e r i a o f s i m i l a r i t y)

\frac{D P}{D E} = \frac{P Q}{E F} [B y d e f . o f s i m i l a r i t y]

\frac{A B}{D E} = \frac{P Q}{E F} [∵ D P = A B]

\Rightarrow P Q = B C

S o i n △ A B C a n d △ D P Q,

A B = D P, A C = D Q a n d B C = P Q

A B = D P, A C = D Q a n d B C = P Q

△ A B C ≅ △ D P Q (b y S S S c r i t e r i o n o f c o n g r u e n c e)

From (i) and (ii), we have

△ A B C ≅ △ D P Q a n d △ D P Q ∥ △ D E F

\Rightarrow △ A B C ∥ △ D E F

THEOREM 8(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.

G I V E N : I n △ A B C a n d △ D E F, ∠ A = ∠ D a n d \frac{A B}{D E} = \frac{A C}{D F}

T O P R O V E : △ A B C ∥ △ D E F

CONSTRUCTION: Take points P and Q on DE and DF respectively,so that

DP = AB and DQ = AC. Join PQ.

P R O O F : I n △ A B C a n d △ D P Q, w e h a v e

A B = D P, ∠ A = ∠ D a n d A C = D Q

△ A B C ≅ △ D P Q (i) . . . . . (b y S A S C r i t e r i o n o f C o n g r u e n c e)

N o w, \frac{A B}{D E} = \frac{A C}{D F}

\Rightarrow \frac{D P}{D E} = \frac{D Q}{D F} [A B = D P a n d A C = D Q]

\Rightarrow P Q ∥ E F [B y t h e c o n v e r s e o f B P T]

∠ D P Q = ∠ E a n d ∠ D Q P = ∠ F . . . . . [C o r r e s p o n d i n g a n g l e s]

D P Q ∥ D F Q

…(ii), (by AAA-criterion of similarity)

From (i) and (ii), we get

△ A B C ≅ △ D P Q a n d △ D P Q ≅ △ D E F

\Rightarrow △ A B C ∥ △ D E F

Areas of similar triangles:

Theorem 9:

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

Corresponding sides

Given: Triangle ABC is similar to triangle PQR.

∠ A = ∠ P

∠ B = ∠ Q

∠ C = ∠ R

\frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{Q R}

T o p r o v e : \frac{A (△ A B C)}{A (△ P Q R)} = {(\frac{A B}{P Q})}^{2} = {(\frac{A C}{P R})}^{2} = {(\frac{B C}{P Q^{}})}^{2}

C o n s t r u c t i o n : D r a w A M ⊥ B C a n d P N ⊥ Q R .

Proof:

A r e a o f t r i a n g l e s = \frac{1}{2} b a s e \times h e i g h t

A (△ A B C) = \frac{1}{2} B C \times A M

A (△ P Q R) = \frac{1}{2} Q R \times P N

\frac{A (△ A B C)}{A (△ P Q R)} = \frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N} = \frac{B C \times A M}{Q R \times P N}

…. (i)

I n △ A B M a n d △ P Q N

∠ B = ∠ Q . . . . . . (G I V E N)

A n d ∠ M = ∠ N (a s b o t h 90, b y c o n s t r u c t i o n)

∴ △ A B M ∥ △ P Q N (b y A A t e s t) (f r o m s t a t e m e n t s A a n d B)

\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A M}{P N} \dots . (i i)

\frac{A (△ A B C)}{A (△ P Q R)} = \frac{B C \times A M}{Q R \times P N} = \frac{B C \times B C}{Q R \times Q R} = {(\frac{B C}{Q R})}^{2}

Similarly we can show for other side too.

\frac{A (△ A B C)}{A (△ P Q R)} = {(\frac{A B}{P Q})}^{2} = {(\frac{B C}{Q R})}^{2} = (\frac{A C}{P R}) 2

Theorem 10:Theorem of three right angles triangles

If a perpendicular is drawn from the vertex of the right angle of a right angle

triangle to the hypotenuse then triangles on both sides of perpendicular are

similar to the whole triangle and each other.

G i v e n : B D ⊥ A C, i n △ A B C ∠ B = 90

T o p r o v e : △ A B C ∥ △ A D B ∥ △ B D C

Proof:

C o n s i d e r ∆ A B C, △ A D B

∠ A = ∠ A . . . . . . . . . (S a m e a n g l e)

∠ A D B = ∠ A B C . . . . . \dots . (b o t h 90)

△ A B C ∥ △ A D B \dots . (1) (b y A A c r i t e r i a)

i n ∆ A B C, △ B D C

∠ C = ∠ C (S a m e a n g l e)

∠ B D C = ∠ A B C . . . (b o t h 90)

△ A B C ∥ △ A D C \dots (2) (b y A A c r i t e r i a)

H e n c e △ A B C ∥ △ A D B ∥ △ B D C . . . . (F r o m 1 a n d 2)

Theorem 11: Pythagoras theorem:

In a right angle triangle,the square of the hypotenuse is equal to the sum of

squares of other two sides.

G i v e n : △ A B C, ∠ B = 90 °

T o p r o v e : A C^{2} = A B^{2} + B C^{2}

C o n s t r u c t i o n : D r a w B D ⊥ A C

P r o o f : A s B D ⊥ A C

△ A B C ∥ △ A D B ∥ △ B D C . . . . (B y t h e o r e m o f 3 r i g h t t r i a n g l e s)

C o n s i d e r △ A B C, △ A D B

C o n s i d e r △ A B C, △ A D B

\frac{A B}{A D} = \frac{B C}{B D} = \frac{A C}{A B}

\Rightarrow \frac{A B}{A D} = \frac{A C}{A B}

A B^{2} = A D \times A C . . . . (1)

N o w c o n s i d e r △ A B C ∥ △ B D C

\frac{A B}{B D} = \frac{B C}{D C} = \frac{A C}{B C}

\Rightarrow \frac{B C}{D C} = \frac{A C}{B C}

B C^{2} = C D \times A D . . . . . . (2)

= A C^{2}

A B^{2} + B C^{2} = C D \times A C + A D \times A C

= A C (C D + A D) . . . .) \dots \dots (T a k i n g A C c o m m o n)

Hence proved.

Theorem 12: Converse of Pythagoras theorem:

In a triangle, if square of one side is equal to the sum of squares of other two
sides, then the angle opposite the first side is a right angle.

G i v e n : A ∆ A B C, A C^{2} = A B^{2} + B C^{2}

T o P r o v e : ∠ B = 90^{\circ}

C o n s t r u c t i o n : C o n s t r u c t ∆ P Q R, P Q = A B, Q R = B C a n d ∠ Q = 90^{\circ}

Proof: In PQR we can apply Pythagoras theorem

S o, P R^{2} = P Q^{2} + Q R^{2}

O r, P R^{2} = A B^{2} + B C^{2} . . . . . . . . . . . . . . . . . . . . (1) (a s b y c o n s t r u c t i o n, P Q = A B, Q R = B C)

B u t A C^{2} = A B^{2} + B C^{2} . . . . . . . . . . . . . . . . . . . . . . . (2) (G i v e n)

So AC = PR (from 1 and 2)

N o w i n ∆ A B C a n d ∆ P Q R

AC = PR

PQ = AB,

QR = BC

∴ ∆ A B C = ∆ P Q R (b y S S S t e s t)

∴ ∠ Q = ∠ B

S o, ∠ B = 90^{\circ}

Proved..

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06 TRIANGLES

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