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TRIANGLES
Similarity of Triangles
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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
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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
AAA Similarity of Triangles
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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
AAA Similarity of Triangles
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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
AAA Similarity of Triangles
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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.
= = Since corresponding
angles are equal, Δ ABC ~ Δ DEF
A
B C
D
E F
P Q
AB
DE
BC
DF
AC
DF
AAA Similarity of Triangles
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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)
DPQ = DEF , so PQ || EF =
(Corollary to basic proportionality theorem)
A
B C
D
P Q
E F
DP
DE
DQ
DF
AAA Similarity of Triangles
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c) AB < DE= (Construction)---------------(i)
Similarly,
= ------(ii)
From (i) and (ii)
= = Since corresponding
angles are equal, Δ ABC ~ Δ DEF
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
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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
AAA Similarity of Triangles
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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
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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
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Proof: In Δ ABC & Δ DPQ
AB = DP and AC = DQ………… (Construction) BAD = EDQ….(given)
So, by SAS congruence
Δ ABC ≅ Δ DPQ
Then,
DPQ = DEF , so PQ || EF =
-----------------(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
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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
= = =
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Construction: Draw altitudes AM and PN of the
triangles.
Proof : Now,
ar (ABC) = BC × AMand
ar (PQR) = QR × PN
Areas of Similar Triangles
A
B C
P
Q RNM
1
2
1
2
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Areas of Similar Triangles
=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)
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Areas of Similar Triangles
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
×
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Areas of Similar Triangles
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
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