the e tutor - similarities of triangles

8/2/2019 The E Tutor - Similarities of Triangles

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TRIANGLES

Similarity of Triangles



A

B C

D

E F

If in two triangles, the corresponding angles are equal, i.e., if the two triangles are equiangular, then

the triangles are similar.

Given:

Two triangles Δ ABC & Δ DEF, Where ∠ A = ∠ D,

∠ B = ∠ E and ∠ C = ∠ F.

To Prove: Δ ABC ~ Δ DEF

AAA Similarity of Triangles



Construction: Let a point P on the line DE and Q onthe line DF such that AB = DP and AC = DQ, we

join PQ.

Proof: Three cases arise:

a) AB < DEb) AB = DE

c) AB > DE

A

B C

D

E F

P Q




a) AB < DEIn Δ ABC & Δ DPQ

AB = DP and AC = DQ………… (Construction)

BAD = EDQ….(given)

So, by SAS congruenceΔ ABC ≅ Δ DPQ

Then,

ABC = DPQ

ABC = DEF………(given)

DPQ = DEF , so PQ || EF =

(Corollary to basic proportionality theorem)

A

B C

D

E F

P Q

DP

DE

DQ

DF




a) AB < DE= (Construction)---------------(i)

Similarly,

= ------(ii)

From (i) and (ii)

= = Since corresponding

angles are equal, Δ ABC ~ Δ DEF

A

B C

D

E F

P Q

AB

DE

BC

DF

AB

DE

AC

DF

AB

DE

BC

DF

AC

DF




b) AB = DEthus P coincides with E

By A.S.A congruence,

Δ ABC ≅ Δ DEF

AB = DE, BC = EF

and AC = DF

Consequently, Q coincides with F.



A

B C

D

E F

P Q

AB

DE

BC

DF

AC

DF




c) AB < DEIn Δ ABC & Δ DPQ

AB = DP and AC = DQ………… (Construction)

BAD = EDQ….(given)

So, by SAS congruenceΔ ABC ≅ Δ DPQ

Then,

ABC = DPQ

ABC = DEF………(given)


(Corollary to basic proportionality theorem)

A

B C

D

P Q

E F

DP

DE

DQ

DF




c) AB < DE= (Construction)---------------(i)

Similarly,

= ------(ii)

From (i) and (ii)



A

B C

D

P Q

E F

Criteria for Similarity of

Triangles

AB

DE

BC

DF

AB

DE

AC

DF

AB

DE

BC

DF

AC

DF



If in two triangles, sides of one triangle are proportional to(i.e., in the same ratio of ) the sides of

the other triangle, then their corresponding angles

are equal and hence the two triangles are similar.

Given: Two Δ ABC and Δ DPQ where

= =

To Prove: ∠ A = ∠ D, ∠ B = ∠ E and∠ C = ∠ F

AB

DE

BC

DF

AC

DF

A

B C

D

E F




Criteria for Similarity of

Triangles

A

B CD

E F

Construction: Two Points P and Q lies on DEand DF respectively, where

DP = AB and DQ = AC and join PQ

Proof:

= and PQ||EF

So, ∠ P = ∠ E and ∠ Q = ∠ F, Thus

= =

So, = =

P Q

DP

PE

DQ

QF

DPDE

DQDF

PQEF

DP

DE

DQ

DF

BC

EF



SAS similarity

If one angle of a triangle is equal to one angle of the

other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given: Two triangles ABC and DEF Where

= and ∠ A = ∠ D

To Prove: Δ ABC ~ Δ DEF

Construction: Two Points P and Q lieson DE and DF respectively, where

DP = AB and DQ = AC

and join PQ

A

B CD

EF

P Q

ABDE

ACDF



Proof: In Δ ABC & Δ DPQ

AB = DP and AC = DQ………… (Construction) BAD = EDQ….(given)

So, by SAS congruence

Δ ABC ≅ Δ DPQ

Then,


-----------------(CBPT)

So, ∠ A = ∠ D,∠ B = ∠ P and

∠ C = ∠ Q

Therefore, by using AAA similarity

Δ ABC ~ Δ DEF

SAS similarity

A

B C

D

EF

P Q



Areas of Similar Triangles

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

the square of the ratio of their corresponding sides.

Given: Two triangles ABC and PQR where

Δ ABC ~ Δ PQR

To Prove:

area of Δ ABC AB 2 AC 2 BC 2

area of Δ PQR PQ PR QR

A

B C

P

Q R

= = =



Construction: Draw altitudes AM and PN of the

triangles.

Proof : Now,

ar (ABC) = BC × AMand

ar (PQR) = QR × PN


A

B C

P

Q RNM

1

2

1

2




=area of Δ ABC

area of Δ PQR

BC × AM1

2

QR × AN1

2

=BC × AM

QR × AN------------------- (i)

A

B C

P

Q RNM

Now, in Δ ABM and Δ PQN,∠ B = ∠ Q ------------------- (Δ ABC ~ Δ PQR)

and ∠ M = ∠ N --------------------------(90°)

So,

Δ ABM ~ Δ PQN------------- (AA similarity criterion)




A

B C

P

Q RNM

Then ,

AMPN

= ABPQ

------------------- (ii)

Δ ABC ~ Δ PQR ----------------------- (Given)

Also,

AB AC BC

PQ PR QR= = ------------------- (iii)

area of Δ ABC

area of Δ PQR

=AB AM

PQ PN

From (i) and (iii) we get

×




A

B C

P

Q RNM

AB AB

PQ PQ×

Or from (ii)

area of Δ ABC

area of Δ PQR= =

AB 2

PQ

Now using (iii), we getarea of Δ ABC

area of Δ PQR=

AB 2

PQ=

AC 2

PR=

BC 2

QR

Proved

the e tutor - similarities of triangles

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