Similarity Of Triangles

Given, ∆DEF ↔ ∆ZXY,

There are mainly two conditions for correspondence DEF ZXY between DEF and XYZ to be a similarity.

⟹The corresponding angles are congruent.

i.e. m∠D = m∠X ∠D ≅ ∠X

m∠E = m∠Y ∠E ≅ ∠Y

m∠F = m∠Z ∠F ≅ ∠Z

⟹The lengths of the corresponding sides are in proportion.

i.e.

Given, ΔPQR and ΔXYZ

P = 2Q and X = 120

If the correspondence PQRYZX is a similarity then

m∠P = m∠X ... (i)

m∠Q = m∠Y ... (ii)

m∠R = m∠Z ... (iii)

From eq. (i) and (ii)

∠P = 120

Because mP = 2mQ

Hence, 2Q = 120

Q = 60

And ∠Q = ∠Y

Hence, ∠Y = 60.

Given, ΔABC and ΔPQR

And AB : PQ = 4 : 5

AC = 6 and QR = 15

The correspondence ABC PQR between ABC and PQR is a similarity.

Hence,

PR =

PR = 7.5

And

BC =

PR = 12

Given, PQR and DEF are similar for the correspondence PQR EDF.

PQ + QR = 15 and DE + DF = 10

PR = 6

If the correspondence PQRDEF is a similarity.

Hence,

And

From these above two equations

EF

EF = 4

Given, In ABC and PQR, ABCQPR is a similarity

The perimeter of ABC is 15 and perimeter of PQR is 27

And BC = 7 and QR = 9

AB + BC + CA = 15

AB + 7 + CA = 15

AB + CA = 8 ...... (i)

PQ + QR + RP = 27

PQ + 9 + RP = 27

PQ + RP = 18 ...... (ii)

Due to similarity

Hence,

PR =

PR = 14.4

Hence,

AC =

AC = 5

Given, XYZ and DEF are similar for the correspondence XYZEDF

And

By similarity law,

Prove that all the isosceles right angled triangles are similar

According to these figures,

In ∆ABC, AB = BC and m∠B = 90

In ∆XYZ, XY = YZ and m∠Y = 90 i.e. ∆ABC and ∆XYZ are isosceles right angled triangles.

To prove: ∆ABC and ∆XYZ are similar triangles.

Proof: ∆ABC and ∆XYZ are both isosceles right angled triangles in which m∠B = m∠Y = 90

Also AB = BC and XY = YZ, then

Given, For ΔABC and ΔXYZ, ABC ↔ XYZ is a similarity

And AC = 3 and XY = 5

Hence,

AB = 4 and BC = 6

Due to similarity of both the triangles,

⟹ YZ = = 7.5

And,

⟹ XZ = = 3.75

(1) False

For equilateral triangles, all the six correspondences are similarity. But in triangles other than equilateral, the measures of all the angles are not same.

(2) True

Congruent triangles are equal with respect to size and shape, while for triangles to be similar, it is sufficient that their shapes are same.

(3) False

Similar triangles are equal in shape but not in size. For triangles to be congruent, they must be equal in both, shape as well as size. Hence, all similar triangles are not congruent.

(4) True

For ∆ABC, the correspondence ABC↔BAC is a similarity.

∴ ∠A≅∠B

∴BC≅AC

Hence, two sides of ABC are congruent and therefore ∆ABC is an isosceles triangle.

(5) Given statement is true.

The triangle ABC and triangle PQR are similar in correspondence ABC↔QRP

∴ mQ = mA = 50

And mP = mC = 30

In ∆PQR,

mP + mQ + mR = 180

30 + 50 + mR = 180

mR = 100

In ∆LMN and ∆XYZ, the correspondence LMN ZYX is a similarity.

∴ mL = mZ = 50

And mN = mX = 40

Hence, mL + mN = 50 + 40 = 90

the correspondence ABC EFD is a similarity in ∆ABC and ∆DEF

∴

Given, ABC PQR is a similarity in ABC and PQR and the perimeter of ABC is 12 and perimeter of a PQR is 20

Hence, AB : PQ = 3 : 5

Given, In ΔABC, a line parallel to intersects and in D and E respectively as shown in fig.

In ABC, a line parallel to BC intersects AB and AC in D and E respectively.

∴ ...... (i)

...... (ii)

…… (iii)

Also, AB = AD + DB and AC = AE + EC

1) AB = AD + DB

AB = 3.6 + 2.4

∴AB = 6

Now from eq (i)

AE =

AE = 2.7

But, AC = AE + EC

AC = 2.7 + 1.8 = 4.5

AC = 4.5

2) AC = AE + EC

4.2 = AE + 3.15

AE = 1.05

From eq (ii)

AD =

AD = 1.55

Then AB = AD + DB

∴ 6.2 = 1.55 + DB

DB = 4.65

3) AC = AE + EC

∴ AC = 6.4 + 8.0

AC = 14.4

∴ From eq (i)

AD =

AD = 9.6

AB = AD + DB

AB = 9.6 + 12

AB = 21.6

4) From eq (ii)

AC =

AC = 13.8

AB = AD + DB

18.4 = 7.2 + DB

DB = 18.4 – 7.2

DB = 11.2

Then, AC = AE + EC

∴ 13.8 = 5.4 + EC

EC = 8.4

5) From eq (iii)

AB =

AB = 6.8

AB = AD + DB

6.8 = AD + 3.4

AD = 3.4

AC = AE + EC

5.1 = AE + 2.55

AE = 2.55

Given,

Given, 2AF = 3FB

∴

The bisector of ∠C intersects AB in F.

BC =

BC = 4.8

Given, XP : PZ = 4 : 5

XY : YZ = 2 : 3

In ∆XYZ, the bisector of angle Y intersects ZX in P.

(1) ∴

XY =

XY = 5.2

(2)

PZ =

PZ = 5.7

ZX = PZ + XP

ZX = 5.7 + 3.8

ZX = 9.5

Given, In ΔABC, the bisector of ∠A intersects in D

TO PROVE :

1)

2)

Proof: In ABC, the bisector of A intersects in D

∴

By applying componendo

∴

By applying invertendo

BD ...... (i)

Similarly,

DC = PROVED.

Given, ABCD is a trapezium such that || . and such that ||

⟹ First draw DB to intersect MN at P.

Proof : AB || CD and MN || AB

∴ MN || CD

Now, MN || AB and M-P-N

Hence, MP || AP

Similarly, as MN || CD and M-P-N

PN || CD

In ∆ABD, MP || AB

... (i)

In ∆BCD, PN || CD

... (ii)

From eq (i) and (ii)

PROVED.

Given, In ABC, D and E are the mid-points of and respectively. and intersect in G. A line m passing through D and parallel to intersects in K

To prove : AC = 4CK

Proof : In ∆ABC, E is the midpoint of AC.

∴ CE = AC

∴ AC = 2CE ...... (i)

In ∆ABC, D is the midpoint of BC.

∴CD = CB

∴ -...... (ii)

In ∆CBE, C-D-B, C-K-E

∴

From eq (ii)

∴

Hence, CE = 2CK

Substituting this into eq (i)

AC = 2 (2CK)

AC = 4CK

Given, In PQR, , such that Q-X-R. A line parallel to and passing through X intersects in Y. A line parallel to px and passing through Y intersects in Z

To prove :

Proof : In PQR, XY || PR

∴ ...... (i)

In PQX, YZ || PX

∴ ...... (ii)

From eq (i) and (ii),

Given, In ABC, and B-X-C. A line passing through X and parallel to intersects in Y. A line passing through X and parallel to intersects in Z as shown in figure.

To prove : CY² = AC × CZ

Proof : In ∆ABC, XY || AB

∴ ...... (i)

In ∆YBC, XZ || BY

∴ ...... (ii)

From eq (i) and (ii)

∴ CY² = AC × CZ

Given, In ABC, the bisector of ∠A intersects in D and the bisector of ∠ADC intersects in E as shown in figure.

To prove: AB × AD × EC = AC × BD × AE

Proof: In ∆ABC, the bisector of ∠A intersects BC in D.

∴ ...... (i)

In ∆ADC, the bisector of ∠ADC intersects AC in E.

∴ ...... (i)

By multiplying eq (i) and (ii)

∴

∴ AB × AD × EC = AC × BD × AE

Given: In ΔABC,

BD = DC [∵ D is mid-point of BC]

AP = PD [∵ P is mid-point of AD]

Construction: Draw a line l from D parallel to AC and let it intersect BQ at S and AB at R.

Draw line m from D parallel to AC intersecting AC at T.

Draw a line from C parallel to BQ.

Extend AD and let it intersect the parallel drawn from C at point E.

Proof:

In ΔADT,

PQ||DT [∵ by construction]

AP = PD [∵ Given]

Hence, by mid-point theorem,

AQ = QT …(1)

In ΔBQC,

BD = DC [∵ given: d is mid-point of BC]

DS|| QC [∵ by construction]

Hence, by mid-point theorem,

QT = TC …(2)

By eq. (1), (2) and (3) we get that,

AQ = QT =TC …(4)

⇒ AQ = 2QC [As QC = QT + TC]

In ΔPDB and ΔCDE,

∠CDE = ∠BDP [vertically opposite angles]

BD = DC [Given]

∠PBD = ∠ECD [alternate interior angles between BP||CE]

Therefore, ΔCDE ≅ ΔPDB.

CE = BP [By CPCT] …(5)

In ΔAPQ and ΔAEC,

∠A = ∠A [common angle]

∠APQ = ∠AEC [Corresponding angles when PQ||CE]

∠AQP = ∠ACE [Corresponding angles when PQ||CE]

By AAA similarity,

ΔAPQ ∼ ΔAEC

And,

[from (4)]

[From (5)]

PQ = 3BP

Hence proved.

In Δ ABC

Given, A-P- B, B- R-S-C and A-Q-C

and

To prove – BR = CS

Proof : In Δ ABC given

∴ ………..eq(1)

(∵ If a line to one side of a Δ intersects the other two sides of the Δ in distinct points, the segments of the other sides of the Δ in the same half plane of the line are proportional to the corresponding sides of the Δ)

Similarly,

∴ ………………..2

From 1 and 2, we get

……..eq(3)

Similarly,

∴ ……………4

From 3 and 4

⇒ BR = CS hence proved

Given ABCD is a ∥gram

To prove : - Δ ABD ∼ Δ BDC

Proof :- For the correspondence ABD ↔ CDB between ΔABD and Δ BDC

∠ BAD ≅ ∠ DCB (opposite ∠s of a ∥gram

∠ABD ≅ ∠CDB (alternate ∠s)

∠ADB ≅ ∠CDB (alternate ∠s)

AB = CD (opposite sides of a ∥ gram ABCD )

∴

Also AD = CB (opposite sides of a ∥ gram ABCD)

∴ = 1

And BD = DB (common)

∴ = 1

Hence, for the correspondence, ABD ↔ CDB between ΔABD and ΔBDC

The corresponding angles are congruent and the lengths of its corresponding sides are in proportion.

∴for the correspondence ABD ↔ CDB, ΔABD and ΔBDC are similar.

Given: In □ ABCD, M and N are the mid points ofand and ∥

To Prove: ∥

Construction:- extend towards D and towards C to intersect each other at P ( ∵ AB > CD)

Proof:- M is the mid-point of

∴ DA = 2 DM …………1

Similarly, N is the mid-point of

∴ CB = 2 CN…………..eq(2)

In Δ PAB , ∥

∴

( ∵ If a line ∥ to one side of a Δ intersects the other two sides of the Δ in distinct points, the segments of the other sides of the Δ in the same half plane of the line are proportional to the corresponding sides of the Δ)

∴ (from 1 and 2)

∴

In Δ PMN, P-D-M , P-C-N and

And

∴ ∥

Given: In □ ABCD, A—P—D, B—Q—C, ∥ and

To prove: AP×QC = PD×BQ

Proof: ∥ and

∴ ∥ ∥

So, , , are three ∥ linesand and are their transversal

∴

(∵ if three or more than three ∥ lines are intercepted by two transversal, the segments cut off on the transversal between the same parallel lines are proportional)

∴ AP × QC = PD × BQ hence proved

Given : In Δ ABC, the bisector of ∠ A intersects in D. The bisector of ∠ADB intersects in F and the bisector of ∠ ADC intersects in E

To prove : AF × AB × CE = AE × AC × BF

Proof : in Δ ABC, the bisector of ∠A intersects at D

∴ 1

(∵ in a Δ the bisector of an angle divides the side opposite to the angle in the segments whose lengths are in the ratio of their corresponding sides)

Similarly, In Δ ADB, the bisector of ∠D intersects at F

∴ ………………2

And in Δ ADC , the bisector of ∠D intersects at E

∴ …………..eq(3)

Multiplying 1, 2 and 3, we get

And
(∵ )

∴

∴ AF × BD × CE =AE × CD × BF

Given:- In Δ ABC, Δ PQR and ΔXYZ correspondences ABC ↔ PQR, PQR ↔ XYZ are similarity

To Prove:- ABC ↔ XYZ is similarity

Proof:- In ΔABC and ΔPQR , the correspondence ABC ↔ PQR is a similarity

∴ ∠A ≅ ∠P, ∠ B ≅ ∠Q , ∠C ≅ ∠R ………..1

…………………..eq(2)

In Δ PQR and ΔXYZ , the correspondence PQR ↔ XYZ is a similarity

∴ ∠ P ≅∠X , ∠ Q ≅ ∠Y , ∠R≅ ∠Z………………3

……………………………………..4

From 1 and 3 , we get

∠A ≅ ∠x, ∠ B ≅ ∠Y , ∠C ≅ ∠Z

Multiplying 2 and 4, we get

Thus for the correspondence ABC ↔ XYZ between ΔABC and ΔXYZ, the corresponding angles are congruent and the lengths of corresponding sides are in proportion.

∴ the correspondence ABC ↔ XYZ between ΔABC and ΔXYZ is a similarity.

(1) The given statement is true because for a line in the plane of a Δ, there are 3 possibilities

i) The line does not intersect the Δ

ii) The line intersects the Δ at one point

iii) The line intersects the Δ at two points

The first possibility is satisfied; hence the given statement is true.

(2) The given statement is false because according to the theorem, if a line lying in the plane of a Δ and not passing through any vertex intersects one side, then it does intersect one more side but does not intersect the third side. Thus, a line not passing through a vertex of a Δ cannot intersect all the three sides.

(3) The given statement is true because a line intersecting a Δ and not passing through any vertex will intersect the Δ at two points.

(4) The given statement is false because if a line intersects two sides of a Δ and does not intersect the third side, it can intersect the line containing the third side.

(5) The given statement is true because in the plane of ΔABC, if line l intersects and at distinct points P and Q respectively, then according to the theorem, if a line lying in the plane of a Δ and not passing through any vertex intersects one side, then it does intersect one more side but does not intersect the third side, it cannot intersect.

In ∆ABC, ∠B is a right angle, AB = 8 and BC = 6

= 24 …(1)

In ∆ABC, ∠B is a right angle and BD is an altitude,

In ∆ABC,

∠A + ∠C = 90⁰ and,

In ∆BDC,

∠DBC + ∠C = 90⁰

So, ∠A = ∠DBC …(2)

In ∆ABC and ∆BDC,

∠DAB ≅ ∠DBC [from (2)]

∠ADB ≅ ∠BDC [by right angles]

The correspondence ADB ↔ BDC is a similarity by AA corollary.

Areas of similar triangles are proportional to the squares of their corresponding sides.

…(3)

Now, ADB + BDC = ABC

BDC = 8.64

The area of ∆BDC is 8.64.

In ABCD, T and intersects in M and in O.

To prove:

AM² = MT •MO

In ∆AMB and ∆OMD

∠AMD ≅ ∠OMD (Vertically opposite angles)

∠MBA ≅ ∠MDO (Alternate angles)

The correspondence AMD↔OMB is a similarity

…(1)

In ∆AMD and ∆TMB

∠AMD ≅ ∠TMB (Vertically opposite angles)

∠MAD ≅ ∠MTB (Alternate angles)

The correspondence AMD↔TMB is a similarity

…(2)

Multiplying (1) and (2),

Hence, AM² = MT •MO

In ABCD, M is the mid-point of. and intersect in N.

To prove : DN = 2MN

Proof: M is the mid-point of BC

MB = MC

…(1)

In ∆MBN and ∆MCD

∠BMN ≅ ∠CMD (Vertically opposite angles)

∠MNB ≅ ∠MDC (Alternate angles)

The correspondence MBN↔MCD is a similarity

(from 1)

MN = MD

Now, D-M-N

DN = DM + MN

DN = DM + MN

DN = MN + MN (MD = MN)

DN = 2MN

Hence proved.

In ∆ABC, P and Q are mid-points of AB and AC respectively.

The correspondence ∆APQ↔∆ABC is a similarity by SSS theorem.

Now,

Areas of similar triangles are proportional to the squares of their corresponding sides.

ABC = 12√3 × 4

ABC = 48√3

The area of ∆ABC is 48√3.

Given: ABCD is a rhombus. = {0}.

To prove: Area of OAB = (area of □ ABCD)

Proof:

O is the mid-point of AC as well as BD.

Further, in rhombus ABCD,

AB ≅ BC ≅ CD ≅ DA

In ∆OAB and ∆OBC,

OA ≅ OC

OB ≅ OB

AB ≅ CB

∆OAB and ∆OBC are congruent by SSS theorem for congruence.

Thus, their areas are equal.

Similarly, ∆OAB, ∆OBC, ∆OCD and ∆ODA are all congruent triangles having equal areas.

∆OAB = ∆OBC = ∆OCD = ∆ODA

Now, ABCD = ∆OAB + ∆OBC + ∆OCD + ∆ODA

ABCD = ∆OAB + ∆OAB + ∆OAB + ∆OAB

ABCD = 4∆OAB

Thus,

Given: In PQRS, PR ∩ QS = {T}, PS = QR and PQ||RS.

To prove: ∆TPS and ∆QTR are similar.

Construction: Through R, draw a line parallel to PS to intersect PQ at M.

Proof:

In PMRS, PM||RS (P-M-Q) and PS||MR (construction).

Thus, PMRS is a parallelogram.

PS = MR

Further, PS = QR (given)

In ∆RMQ, ∠RMQ ≅ ∠RQM …(i)

Further, the corresponding angles formed by transversal PQ to PS||MR are congruent.

∠RMQ ≅ ∠SPM (corresponding angles)

∠RQM ≅ ∠SPM (by (i))

∠RQP ≅ ∠SPQ (P-M-Q)

In ∆SPQ and ∆RQP,

∠SPQ ≅ ∠RQP

PQ ≅ QP (same line segment)

SP ≅ RQ (given)

The correspondence ∆SPQ↔∆RQP is a congruence by SAS theorem of congruence.

∠PSQ ≅ ∠QRP

∠PST ≅ ∠QRT (S-T-Q and R-T-P)

In ∆TPS and ∆QTR

∠STP ≅ ∠RTQ (Vertically opposite angles)

∠PST ≅ ∠QRT

∆TPS and ∆QTR are similar by AA corollary.

Thus, by AA corollary the correspondence ∆TSP↔∆TRQ is a similarity.

Given: In ∆ABC, a line parallel to BC, passes through the mid-point of AB.

To prove: Line m bisects AC.

Proof: In the plane of ∆ABC, line m is parallel to BC and intersects AB at a point other than a vertex of the triangle.

Thus, m intersects AC.

Let m ∩ AC = {E}

In ∆ABC, D is the mid-point of AB.

AD = DB

In ∆ABC, A-D-B, A-E-C and DE||BC.

AE = EC and A-E-C.

E is the mid-point of AC.

Thus, line m bisects AC.

In ∆ABC, P, Q, R are the mid-points of the sides AB, BC and CA respectively.

The correspondence ∆ABC↔∆QRP is a similarity.

Areas of similar triangles are proportional to the squares of their corresponding sides.

∆ABC = ∆POR

Similarly, X,Y and Z are the mid-points of the sides of ∆PQR, we get

∆PQR = 4∆XYZ

∆PQR = 4×10

∆PQR = 40

Thus, the area of ∆PQR is 40 sq. units.

∆ABC = 4∆PQR

∆ABC = 4×40

∆ABC = 160

Hence, the area of ∆ABC is 160 sq. units.

Given: In ∆ABC, X∊BC, Y∊CA, Z∊AB.

Also, XY||AB, YZ||BC and ZX||AC.

To prove: X,Y and Z are the mid-points of BC, CA and AB respectively.

Proof: In ∆ABC, YZ||BC.

…(1)

In ∆ABC, XY||AB.

…(2)

From (1) and (2),

…(3)

In ∆ABC, ZX||AC.

…(4)

From (3) and (4),

AZ² = BZ²

AZ = BZ

Hence, A-Z-B and AZ = BZ.

Thus, Z is the mid-point of AB.

Similarly, Y is the mid-point of AC and X is the midpoint of BC.

Therefore, X, Y and Z are the mid-points of BC, CA and AB respectively.

Given: The correspondence ∆ABC↔∆PQR is a similarity and AB ≅ PQ.

To prove: ∆ABC and ∆PQR are congruent.

Proof:

The correspondence ∆ABC↔∆PQR is a similarity.

Further, AB ≅ PQ

AB = PQ

AB = PQ, BC = QR, AC = PR

AB ≅ PQ, BC ≅ QR, AC ≅ PR

The correspondence ∆ABC↔∆PQR is a congruence by SSS theorem of congruence.

Hence, ∆ABC and ∆PQR are congruent.

We have

Given: In ∆ABC, such that , where m and n are positive real numbers.

QP ∥ BC

Prove that: (m + n)² (area of ∆APB) = m² (area of ∆ABC)

Proof: Since

[by invertendo componendo invertendo rule]

⇒

In ∆APQ and ∆ABC,

∠APQ ≅ ∠ABC [∵, corresponding angles]

∠AQP ≅ ∠ACB [∵, corresponding angles]

⇒ The correspondence APQ ↔ ABC is a similarity.

Remember the property, areas of similar triangles are proportional to the squares of the corresponding sides.

∴

⇒

⇒

⇒

By cross-multiplication, we get

(m + n)² (Area of ∆APB) = m² (Area of ∆ABC)

Thus, proved.

We have

Given: D is the midpoint of , E is the midpoint of & F is the midpoint of .

⇒ AF = FB, AE = EC & BD = DC

To Prove: Area of □BDEF = 1/2 Area of ∆ABC

Proof: Since, given is that D, E and F are midpoints of respectively.

We know, AE = EC.

⇒ AE = FD [∵, EC = FD]

⇒ [∵, ] …(i)

Also, we know that, AF = FB.

⇒ AF = ED [∵, FB = ED]

⇒ [∵, ] …(ii)

Now, in ∆AFE and ∆DEF, we have

AF ≅ ED, AE ≅ FD & FE ≅ EF

⇒ By SSS theorem of congruence, we can say that the congruency AFE ↔ DEF is a similarity.

Similarly, in ∆AFE and ∆FBD, we have

AF ≅ FB, AE ≅ FD & FE ≅ BD

⇒ By SSS theorem of congruence, we can say that the congruency AFE ↔ FBD is a similarity.

Similarly, the correspondence AEF ↔ EDC is a similarity.

So, ∆AFE, ∆DEF, ∆FBD and ∆EDC are all congruent triangles.

⇒ ∆AFE = ∆DEF = ∆FBD = ∆EDC …(iii)

Now,

BDEF = ∆DEF + ∆FBD

⇒ BDEF = ∆AFE + ∆AFE [from equation (iii)]

⇒ BDEF = 2 ∆AFE …(iv)

And

∆ABC = ∆AFE + ∆DEF + ∆FBD + ∆EDC

⇒ ∆ABC = ∆AFE + ∆AFE + ∆AFE + ∆AFE [from equation (iii)]

⇒ ∆ABC = 4 ∆AFE

⇒ ∆ABC = 2 (2 ∆AFE) …(v)

Comparing equations (iv) and (v), we get

∆ABC = 2 (BDEF)

⇒ BDEF = 1/2 ∆ABC

⇒ Area of □BDEF = 1/2 (Area of ∆ABC)

Thus, proved.

Yes, two similar triangles can have same area.

Note, when two triangles are congruent then all corresponding sides as well as corresponding angles of one triangle are equal to those of the other triangle.

⇒ If two triangles are congruent, they are similar too.

And congruent triangles have same area too.

Hence, the similar triangles need to be congruent to have similar areas.

We have

Given: The correspondence ABC ↔ DEF is similarity in ∆ABC and ∆DEF.

is an altitude of ∆ABC and is an altitude of ∆DEF.

To Prove: AB × DN = AM × DE

Proof: Since, the correspondence ABC ↔ DEF is a similarity.

By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

So, we can write ∠B ≅ ∠E. …(i)

In ∆ABM and ∆DEN,

From result (i),

∠B ≅ ∠E

Also, ∠M ≅ ∠N [since, they are right angles of ∆ABM and ∆DEN respectively]

⇒ By AA-corollary, the correspondence ABM ↔ DEN is a similarity in ∆ABC and ∆DEF]

Then, again by definition of similarity of correspondences in triangles we can say that,

By cross-multiplication, we get

⇒ AB × DN = AM × DE

Hence, proved.

(1). Similarity of triangles and congruence of triangles are closely related, but has a fine line of difference.

Similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length.

By AAA criterion of similarity of triangles, we get the shapes of triangles equal but, for congruence of triangles their sizes must be equal too.

∴ AAA criterion of similarity of triangles cannot be the criterion for congruence of triangles.

Hence, the statement is true.

(2). For congruence of triangles, the corresponding sides are of equal length while for similarity of triangles, measure of the corresponding sides of two triangles must be proportional.

SAS criterion for congruence of triangles not only makes sure that the triangles have equal length of corresponding sides but also makes sure that the triangles have same shape.

∴ SAS criterion for congruence of triangles can be a criterion for similarity of triangles.

Hence, the statement is false.

(3). Congruency of triangles ensure that the triangles have same shape as well as equal lengths of corresponding sides.

⇒ The shape and size of two congruent triangles are equal.

∴ Two congruent triangles have the same area.

Hence, the statement is true.

(4). Similarity of triangles indicate that they have the same shape but their sizes are not same.

Area of triangles are equal when their size are equal, shape of triangle cannot determine its area.

∴ Two similar triangles doesn’t always have the same area.

Hence, the statement is false.

(5). Area of similar triangles are proportional to the squares of their corresponding sides.

Since, the measure of corresponding angles of similar triangles are always equal.

∴ Area of similar triangles are not proportional to the squares of their corresponding angles.

Hence, the statement is false.

We have

Given that: AD and BE are altitudes of ∆ABC. ⇒ ∠ADC = ∠

In ∆ADC and ∆BEC,

∠ADC = ∠BEC [∵, ∠ADC = ∠BEC = 90° as AD and BE ]

∠ACD = ∠BCE [∵, ∠ACD and ∠BCE are same angles of the same triangles, so they obviously are equal]

⇒ By AA-corollary, we can say that ∆ADC ∼ ∆BEC for the correspondence of ADC ↔ BEC.

By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒

⇒ BC = 8.8

So, perimeter of ∆ABC is given by

Perimeter of ∆ABC = AB + BC + AC

⇒ Perimeter of ∆ABC = 12 + 8.8 + 9.9

⇒ Perimeter of ∆ABC = 30.7

Thus, perimeter of ∆ABC = 30.7.

We have

Given: D is the midpoint of PQ ⇒ PD = DQ = 1/2 PQ

But PD = DQ = FE [∵, F and E are also midpoints of PR and QR respectively]

So, EF = 1/2 PQ

Or …(i)

E is the midpoint of QR ⇒ QE = ER = 1/2 RQ

But QE = ER = DF [∵, D and F are also midpoints of PQ and PR respectively]

So, DF = 1/2 RQ

Or …(ii)

F is the midpoint of PR ⇒ PF = FR = 1/2 RP

But PF = FR = DE [∵, D and E are midpoints of PQ and QR respectively]

So, DE = 1/2 RP

Or …(iii)

Comparing equations (i), (ii) & (iii), we get

We can also arrange it as,

∴ The correspondence DEF ↔ RPQ is a similarity by SSS theorem.

Thus, answer is RPQ.

We have

Given: AM ⊥ BC and CN ⊥ AB.

AB = 12, BC = 15 & AM = 9.6

To find: CN = ?

For this,

Take ∆AMB and ∆CNB,

∠AMB = ∠CNB [∵, ∠AMB = ∠CNB = 90°]

∠ABM = ∠CBN [∵, ∠AMB and ∠CBN are same angles of the same triangle ABC]

⇒ By AA corollary, ∆AMB ∼ ∆CNB for the correspondence AMB ↔ CNB.

By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

Or

⇒

⇒

⇒

⇒ CN = 12

Thus, answer is 12.

Given: Area of two similar triangles.

Area of ∆₁ = 25

Area of ∆₂ = 16

Recall two properties:

First,

Ratio of areas of two similar triangles = Ratio of the squares of the corresponding sides

Second,

Ratio of perimeter of two similar triangles = Ratio of corresponding sides

Using the first property, we can write that

⇒

Taking under root of both sides, we get

⇒

⇒

Now, using second property, we can say

Thus, answer is 5/4.

Given: Area of ∆ABC = 36,

Area of ∆PQR = 64

AB = 12

The correspondence ABC ↔ PQR is a similarity.

To find: PQ = ?

For this,

Recall the property,

Ratio of areas of two similar triangles = Ratio of squares of the corresponding sides

⇒

⇒

⇒

⇒

⇒

⇒

⇒ PQ = 2 × 8

⇒ PQ = 16

Thus, answer is 16.

We have

Given: A – M – B, A – N – C

MN ∥ BC

AM = 8.4

AN = 6.4

CN = 19.2

To find: AB = ?

Take ∆ABC,

We know that, MN ∥ BC. So, by property we can write as:

⇒

⇒

⇒

⇒ BM = 25.2

We know AM = 8.4 and BM = 25.2.

Now, AB = AM + BM [∵, from the diagram]

⇒ AB = 8.4 + 25.2

⇒ AB = 33.6

Thus, the answer is 33.6.

Given: The correspondence ABC ↔ PQR is a similarity in ∆ABC and ∆PQR.

AB = 16,

AC = 8,

PQ = 24 &

BC = 12

To find: QR + PR = ?

By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒

To find (QR + PR), cross multiply it.

⇒ (QR + PR) × AB = (BC + AC) × PQ

Substituting values,

⇒ (QR + PR) × 16 = (12 + 8) × 24

⇒ (QR + PR) × 16 = 20 × 24

⇒

⇒ QR + PR = 30

Thus, the answer is 30.

Given: The correspondence ABC ↔ XYZ is a similarity in ∆ABC and ∆XYZ.

ABC = 72,

BC = 6 &

YZ = 10

To find: XYZ = ?

So, by property we can say that,

Ratio of areas of two similar triangles = Ratio of squares of the corresponding sides

⇒

⇒

Taking reciprocal of both sides, we get

Substituting values in the above equation, we get

⇒

⇒

⇒

⇒ XYZ = 200

Thus, the answer is 200.

We have

Given: ABCD is a trapezium.

AD ∥ BC

PD = 9,

AP = 5 &

PB = 7.2

To find: AC = ?

We have got the trapezium ABCD, in which

AD ∥ BC & P is the intersection point of AC and BD.

Take ∆PAD and ∆PCB,

∠APD = ∠CPB [∵, Vertically opposite angles are equal]

∠PAD = ∠PCB [∵, alternate angles]

∠PDA = ∠PBC [∵, alternate angles]

⇒ By AAA similarity theorem, ∆PAD ∼ ∆PCB for the correspondence PAD ↔ PCB.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

Reciprocating it, we get

⇒

⇒

⇒ PC = 4

From the diagram, note that

AC = AP + PC

⇒ AC = 5 + 4

⇒ AC = 9

Thus, the answer is 9.

We have

Given: In ∆ABC,

∠ABC = ∠ADB = ∠BDC = 90°

In ∆BDA and ∆CDB,

∠BDA = ∠CDB = 90°

∠ABD = ∠BCD [measure of both being equal, i.e., (90° - m∠A)]

This can be explained as,

In ∆BDA, by angle sum property of triangle,

∠BDA + ∠DAB + ∠ABD = 180°

⇒ 90° + ∠A + ∠ABD = 180° [∵, ∠BDA = 90° & ∠DAB = ∠A]

⇒ ∠ABD = 180° - 90° - ∠A

⇒ ∠ABD = 90° - ∠A …(i)

Similarly, in ∆ABC by angle sum property of triangle,

∠ABC + ∠BAC + ∠BCA = 180°

⇒ 90° + ∠A + ∠BCD = 180° [∵, ∠ABC = 90°, ∠BAC = ∠A & ∠BCA = ∠BCD (from the figure)]

⇒ ∠BCD = 180° - 90° - ∠A

⇒ ∠BCD = 90° - ∠A …(ii)

By equations (i) and (ii), we get

∠ABD = ∠BCD

Hence, by AA corollary the correspondence BDA ↔ CDB is similarity between ∆BDA and ∆BDC.

Thus, the answer is CDB.

Given: The correspondence ABC ↔ ZXY is a similarity in ∆ABC and ∆XYZ.

AB = 12,

BC = 9,

CA = 7.5 &

ZX = 10

To find: YZ + XY = ?

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒

Reciprocating both sides,

⇒

⇒

⇒

⇒

Thus, the answer is 13.75.

Given: m ∠A = 40°

The correspondence ABC ↔ DEF is a similarity between ∆ABC and ∆DEF.

To find: m ∠E + m ∠F = ?

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒ m ∠A = m ∠D

⇒ m ∠A = m ∠D = 40°

⇒ m ∠D = 40°

In ∆DEF,

m ∠D + m ∠E + m ∠F = 180°

⇒ 40° + m ∠E + m ∠F = 180°

⇒ m ∠E + m ∠F = 180° - 40°

⇒ m ∠E + m ∠F = 140°

∴ Answer is 140.

Thus, option (c) is correct.

We have the diagram as

Given: In ∆ABC, MN ∥ BC.

By property of triangles, we can say that

,

&

Cross multiplication of the above equations,

⇒ AM × NC = AN × MB

Or AM . NC = AN . MB is true.

AM × AC = AN × AB

Or AM . AC = AN . AB is true.

MB × AC = NC × AB

Or MB . AC = NC . AB is true.

So, AM . MB = AN . NC is not true.

Thus, option (b) is correct.

We have the diagram as,

Given: In ∆ABC,

MN ∥ AB

NC:NA = 1:3

⇒

CM = 4

To find: BC = ?

In ∆ABC,

Since, MN ∥ AB

⇒

⇒

⇒ MB = 4 × 3 [∵, ⇒ ]

⇒ MB = 12

In B – M – C,

BC = MB + CM

⇒ BC = 12 + 4

⇒ BC = 16

Thus, option (b) is correct.

For ∆XYZ and ∆PQR, the correspondence XYZ ↔ PQR has similarity.

Also, given that

XY = 12,

YZ = 8,

ZX = 16 &

PR = 8

To find: PQ + QR = ?

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

Or

⇒

⇒

⇒

⇒

⇒ PQ + QR = 10

Thus, option (b) is correct.

We have

Given: PD is the bisector of ∠P.

QD:RD = 4:7

PR = 14

To find: PQ = ?

In ∆PQR, the bisector of P intersects the line QR at D. By property of triangles,

⇒

⇒

⇒

⇒ PQ = 2 × 4

⇒ PQ = 8

Thus, option (a) is correct.

Given: In ∆ABC,

BC:CA:AB = 3:4:5

From ∆ABC and ∆PQR, the correspondence ABC ↔ PQR is similarity.

In ∆PQR,

PR = 12

To find: Perimeter of ∆PQR = ?

In ∆ABC, since we have BC:CA:AB = 3:4:5

Let the lengths of BC, CA and AB be 3t, 4t and 5t respectively. (where, t > 0)

Perimeter of ∆ABC = 3t + 4t + 5t

⇒ Perimeter of ∆ABC = 12t, t > 0

Also, from given we have,

In ∆ABC and ∆PQR, the correspondence ABC ↔ PQR is similarity.

So, from property we can say that,

Ratio of perimeters of ∆ABC and ∆PQR = Ratio of their corresponding sides

⇒

Substituting the given values, we get

⇒

⇒ Perimeter of ∆PQR = 36

Thus, option (b) is correct.

Given that,

From ∆ABC and ∆XYZ, the correspondence ABC ↔ YZX is similarity.

Also, m ∠B + m ∠C = 120°

To find: m ∠Y = ?

In ∆ABC, by angle sum property of triangles we can say that,

m ∠A + m ∠B + m ∠C = 180°

⇒ m ∠A + (m ∠B + m ∠C) = 180°

⇒ m ∠A + 120° = 180°

⇒ m ∠A = 180° - 120°

⇒ m ∠A = 60° …(i)

Now, from ∆ABC and ∆XYZ, the correspondence ABC ↔ YZX is similarity.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒ ∠A ≅ ∠Y

∴ m ∠A = m ∠Y

From equation (i), we get

m ∠A = m ∠Y = 60°

∴ m ∠Y = 60°

Thus, option (d) is correct.

Given: AB + BC = 10,

DE + EF = 12 &

AC = 6

The correspondence ABC ↔ DEF is similarity between ∆ABC and ∆DEF.

To find: DF = ?

From ∆ABC and ∆DEF, the correspondence ABC ↔ DEF is a similarity.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒

By reciprocating on both sides,

⇒

⇒

⇒

⇒ DF = 7.2

Thus, option (c) is correct.

Given: Lengths of sides of ∆DEF are 4, 6 and 8.

∆DEF ∼ ∆PQR for the correspondence DEF ↔ PQR.

Perimeter of ∆PQR = 36

To find: length of the smallest side of ∆PQR = ?

Since, sides of ∆DEF are 4, 6 and 8.

⇒ Perimeter of ∆DEF = 4 + 6 + 8

⇒ Perimeter of ∆DEF = 18

Since, ∆DEF ∼ ∆PQR for the correspondence DEF ↔ PQR.

∴

⇒

⇒

⇒

⇒ Smallest side of ∆PQR = 4 × 2

⇒ Smallest side of ∆PQR = 8

Thus, option (d) is correct.

We have

Given: ∠ABD = ∠CBD

BA = 12,

BC = 16 &

AD = 9

To find: AC = ?

∵, In ∆ABC bisector of ∠B intersects AC at D.

By theorem of proportionality, we have

By reciprocating on both sides,

⇒

Substituting the values, we get

⇒

⇒

⇒ CD = 12

We can express AC as,

AC = AD + CD

⇒ AC = 9 + 12

⇒ AC = 21

Thus, option (b) is correct.

We have,

Given: In ∆ABC,

PQ ∥ BC,

4 AP = AB &

AQ = 4

To find: AC = ?

We know, 4 AP = AB

⇒ …(i)

In ∆ABC, we know that PQ ∥ BC.

⇒

⇒

⇒

⇒ AC = 4 × 4

⇒ AC = 16

Thus, the option (d) is correct.

Given: In ∆ABC, the correspondence ABC ↔ BAC is similarity.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

∠A ≅ ∠B, ∠B ≅ ∠A and ∠C ≅ ∠C are all true.

Thus, option (c) is true.

We have

In ∆ABC,

PQ ∥ BC

PQ = 5,

AP = 4 &

AB = 12

In ∆ABC and ∆APQ, we have

∠A = ∠A [∵, ∠BAC and ∠PAQ are equal angles as they are same angles]

∠ABC = ∠APQ [∵, Alternate angles are equal]

∠ACB = ∠AQP [∵, Alternate angles are equal]

⇒ By AAA-corollary of similarity, ∆ABC ∼ ∆APQ for the correspondence ABC ↔ APQ.

Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.

⇒

⇒

⇒ BC = 5 × 3

⇒ BC = 15

Thus, option (c) is correct.

We have

Given: P – M – Q and P – N – R

In ∆PQR, MN ∥ QR.

PQ = 18,

PM = 12 &

PR = 9

To find: NR = ?

In ∆PQR, when MN ∥ QR.

⇒

Substituting the given values in the above equation,

⇒

⇒ PN = 6

Observe the diagram,

PR = PN + NR

⇒ NR = PR – PN

⇒ NR = 9 – 6

⇒ NR = 3

Thus, option (b) is correct.