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The ∆ ABC ≅ ∆ FED then

The corresponding vertices A and F, B and E, C and D. The corresponding sides are \(\bar{AB}\) and FE, BC and ED, CA and DF.

The corresponding angles are ∠A and ∠F, ∠B and ∠E and ∠C and ∠D.

ΔCAB ≅ ΔQRP

Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal.

If we write ΔCAB ≅ ΔQRP, it would mean that,

Correspondence between vertices:

C ↔ Q, A ↔ R, B ↔ P

Correspondence between sides:

CA = QR, AB = RP, BC = PQ

Correspondence between angles:

∠C = ∠Q, ∠A = ∠R, ∠B = ∠P

The statement is false.

If two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle then the two triangles must be congruent.

From the figure we know that D is the midpoint of the line AC

So we get

AD = CD = ½ AC

It is given that BD = ½ AC

So we can write it as

AD = BD = CD

Let us consider AD = BD

We know that the angles opposite to equal sides are equal

So we get

∠ BAD = ∠ ABD …..(1)

Let us consider CD = BD

We know that the angles opposite to equal sides are equal

So we get

∠ BCD = ∠ CBD ….. (2)

By considering the angle sum property in △ ABC

We get

∠ ABC + ∠ BAC + ∠ BCA = 180^o

So we can write it as

∠ ABC + ∠ BAD + ∠ BCD = 180^o

By using equation (1) and (2) we get

∠ ABC + ∠ ABD + ∠ CBD = 180^o

So we get

∠ ABC + ∠ ABC = 180^o

By addition

2 ∠ABC = 180^o

By division

∠ABC = 90^o

Therefore, ∠ABC is a right angle.

Based on the △ AOD and △ BOC

From the figure we know that ∠ AOD and ∠ BOC are vertically opposite angles.

So we get

∠ AOD = ∠ BOC

We also know that

∠ DAO = ∠ CBO = 90^o

It is given that AD = BC

Therefore, by AAS congruence criterion we get

△ AOD ≅ △ BOC

So we get AO = BO (c. p. c. t)

Therefore, it is proved that CD bisects AB.

(i) From the figure △ AOB and △ DOC

We know that AB || CD and ∠ BAO and ∠ CDO are alternate angles

So we get

∠ BAO = ∠ CDO

From the figure we also know that O is the midpoint of the line AD

We can write it as AO = DO

According to the figure we know that ∠ AOB and ∠ DOC are vertically opposite angles.

So we get ∠ AOB = ∠ DOC

Therefore, by ASA congruence criterion we get

△ AOB ≅ △ DOC

(ii) We know that △ AOB ≅ △ DOC

So we can write it as

BO = CO (c. p. c. t)

Therefore, it is proved that O is the midpoint of BC.

Given that ABC is a right angled triangle

Such that,

∠A = 90°

And,

AB = AC

Since,

AB = AC

ΔABC is also isosceles triangle

Therefore, we can say that ΔABC is a right angled isosceles triangle.

∠C = ∠B

And,

∠A = 90°

Now, we have

Sum of angles in a triangle = 180°

∠A + ∠B + ∠C = 180°

90° + ∠B + ∠B = 180°(From i)

90° + 2∠B = 180°

2∠B = 90°

∠B = 45°

Therefore, ∠B = ∠C = 45°

According to the figure we need to prove that AT = BT

We know that

Angle of incidence = Angle of reflection

So we get

∠ ACN = ∠ DCN ….. (1)

We know that AB || CN and AC is the transversal

From the figure we know that ∠ TAC and ∠ ACN are alternate angles

∠ TAC = ∠ CAN …… (2)

We know that AB || CN and BD is the transversal

From the figure we know that ∠ TBC and ∠ DCN are corresponding angles

∠ TBC = ∠ DCN ….. (3)

By considering the equation (1), (2) and (3)

We get

∠ TAC = ∠ TBC …… (4)

Now in △ ACT and △ BCT

∠ ATC = ∠ BTC = 90^o

CT is common i.e. CT = CT

By AAS congruence criterion

△ ACT ≅ △ BCT

AT = BT (c. p. c. t)

Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.

(i) In ∆ADB and ∆ADC

AD = AD (common)

AB = AC (given)

and DB = DC (∵ D is mid point of BD)

(ii) By S.S.S. rule of congruency.

∆ADB ≅ ∆ADC

(iii) ∠B = ∠C (CPCT)

(i) In ∆ABD and ∆ACD

AB = AC, BD = CD and AD = AD (common)
∴ S.S.S. condition
∆ADB ≅ ∆ADC

(ii) In ∆PQR and ∆PSR

QR = RS, PQ = PS and RP = RP (common)
By S.S.S. condition
∆PQR ≅ ∆PSR

∆NOL ≅ ∆LMN because

∵ OL = NM, ON= LM and NL = NL (common)

∴ By S.S.S. condition,

∆NOL ≅ ∆LMN

∆ABC is not congruent to ∆DEF because in RHS rule, one side and hypotenuse of one right angled triangle are equal to corresponding side and hypotenuse which is not in ∆ABC and ∆DEF

In right angle triangle ∆PRQ and ∆QSP

PR = SQ = 2 cm

PQ = PQ (common)

∠PRQ = ∠QSP = 90° (right angle)

∴ From R.H.S. congruency

∆PRQ ≅ ∆QSP

(i) ∠N is corresponding angle to ∠Z.

(ii) Side LM is corresponding to side AT,

(iii) ∠M is corresponding angle to ∠Y.

(iv) Side MN is corresponding to side YZ.

(i) Three equal parts in ∆CBD and ∆BCE are

CB = BC (common)

∠CDB = ∠BEC (each 90°)

and BD = CE (given)

(ii) From R.H.S. condition of congruency

∆CBD ≅ ∆BCE

(iii) Yes, ∠DCB = ∠EBC

(∵ corresponding angles of congruent triangles are equal)

Two circles will be congruent if radii are equal.

Such figures which covers completely to each other are called congruent.

(i) Length

(ii) sides

(iii) 60°

(i) measure (size)

(ii) congruent

(iii) A ≅ B,

(iv) 6

(i) Our parents or brother or sister have similar mobiles which are same in size and shape.

(ii) We see 100 – 100 rupee notes in purse of our father which are same in size and shape.

Corresponding parts of congurent triangles are congruent.

(i) Duplicate key of a lock.

(ii) ₹500 – 500 note

(iii) Pages of a book.

∆ABC ≅ ∆FED means that ∆ABC covers completely and vertices of ∆ABC will be on he vertices of ∆FED.

∴ A ↔ F, B ↔ E and C ↔ D

Corresponding sides will be congruent.

\(\overline {AB}\) ↔ \(\overline {FE}\) , \(\overline {BC}\) ↔ \(\overline {ED}\) and \(\overline {CA}\) ↔ \(\overline {DF}\)

Corresponding angles will be congruent.

∠A ↔ ∠F, ∠B ↔ ∠E and ∠C ↔ ∠D

It is given that ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP

Consider QR intersecting the line PC at point M

We know that △ ABC is an equilateral triangle

So we get ∠ A = ∠ ACB = 60^o

From the figure we know that PQ || AC and ∠ BPQ and ∠ ACB are corresponding angles

So we get

∠ BPQ = ∠ ACB = 60^o

Based on the △ BPQ we know that

∠ B = ∠ ACB = 60^o

It can be written as

∠ BQP = 60^o

According to the figure we know that △ BPQ is an equilateral triangle

So we get

PQ = BP = BQ

It is given that CR = BP so we get

PQ = CR ……. (1)

In the △ PMQ and △ CMR we know that PQ || AC and QR is the transversal

We know that ∠ PQM and ∠ CRM are alternate angles and ∠ PMQ and ∠ CMR are vertically opposite angles

∠ PQM = ∠ CRM

∠ PMQ = ∠ CMR

By considering equation (1) and AAS congruence criterion

△ PMQ ≅ △ CMR

We know that the corresponding parts of congruent triangles are equal

PM = MC

Therefore, it is proved that QR bisects PC.

Given that

A, B, C of a triangle ABC are equal to each other.

We have to prove that,

Δ ABC is equilateral.

We have,

∠A = ∠B = ∠C

Now,

∠A = ∠B

BC = AC (Opposite sides to equal angles are equal)

∠B = ∠C

AC = AB (Opposite sides to equal angles are equal)

From the above, we get

AB = BC = AC

Therefore, ΔABC is an quadrilateral triangle.

Hence, proved

Given to prove that each angle of the equilateral triangle is 60°

Let us consider an equilateral triangle ABC

Such that,

AB = BC = CA

Now,

AB = BC

∠A = ∠C [i] (Opposite angles to equal sides are equal)

BC = AC

∠B = ∠A [ii] (Opposite angles to equal sides are equal)

From [i] and [ii], we get

∠A = ∠B = ∠C [iii]

We know that,

Sum of all angles of triangles = 180°

∠A + ∠B + ∠C = 180°

∠A + ∠A + ∠A = 180°

3∠A = 180°

∠A = \(\frac{180}{3}\)

= 60°

Therefore, ∠A = ∠B = ∠C = 60°

Hence, each angle of an equilateral triangle is 60°

Yes, the given triangles are congruent as AB = AD, BC = CD and AC is a common side. Hence, by SSS theorem triangle ABC is congruent to triangle ADC.

In ΔABC and ΔA'B'C',

AB = A'B' (Given)

∠B = ∠B'(Given)

BC = B'C' (Given)

Hence, by S.A.S. theorem,

ΔABC ≅ ΔA'B'C'

Therefore,

By c.p.c.t

∠A = ∠A'

3x = 2x +20

x = 20°

Therefore angle A' = 2x +20

= 60°

Since the two triangles with two adjacent sides and an angle adjacent to any one side among them are shown equal, then the two triangles will be similar but not necessarily congruent.

(i) Considering △ APB and △ AQB

We know that

∠ APB = ∠ AQC = 90^o

From the figure we know that l is the bisector of ∠ A

So we get

∠ BAP = ∠ BAQ

We know that AB is common i.e. AB = AB

Therefore, by AAS congruence criterion we get

△ APB ≅ △ AQB

(ii) We know that △ APB ≅ △ AQB

So it is proved that BP = BQ (c. p. c. t)

(i) Based on the △ ABD and △ ACD

From △ ABC we know that AB and AC are equal sides of isosceles triangle

So we get

AB = AC

From △ DBC we know that DB and DC are equal sides of isosceles triangle

So we get

DB = DC

We also know that AD is common i.e. AD = AD

Therefore, by SSS congruence criterion we get

△ ABD ≅ △ ACD

(ii) We know that △ ABD ≅ △ ACD

We get ∠ BAD = ∠ CAD (c. p. c. t)

It can be written as

∠ BAE = ∠ CAE ……… (1)

Considering △ ABE and △ ACE

We know that AB and AC are the equal sides of isosceles △ ABC

AB = AC

So by using equation (1) we get

∠ BAE = ∠ CAE

We know that AE is common i.e. AE = AE

Therefore, by SAS congruence criterion we get

△ ABE ≅ △ ACE

(iii) We know that △ ABD ≅ △ ACD

We get ∠ BAD = ∠ CAD (c. p. c. t)

It can be written as

∠ BAE = ∠ CAE

Therefore, it is proved that AE bisects ∠ A.

Considering △ BDE and △ CDE

We know that BD and CD are equal sides of isosceles △ ABC

Since △ ABE ≅ △ ACE

BE = CE (c. p. c. t)

We know that DE is common i.e. DE = DE

Therefore, by SSS congruence criterion we get

△ BDE ≅ △ CDE

We know that ∠ BDE = ∠ CDE (c. p. c. t)

So DE bisects ∠ D which means that AE bisects ∠ D

Hence it is proved that AE bisects ∠ A as well as ∠ D.

(iv) We know that △ BDE ≅ △ CDE

So we get

BE = CE and ∠ BED = ∠ CED (c. p. c. t)

From the figure we know that ∠ BED and ∠ CED form a linear pair of angles

So we get

∠ BED = ∠ CED = 90^o

We know that DE is the perpendicular bisector of BC

Therefore, it is proved that AE is the perpendicular bisector of BC.

It is given that x = y and AB = CB

By considering the △ ABE

We know that

Exterior ∠ AEB = ∠ EBA + ∠ BAE

By substituting ∠ AEB as y we get

y = ∠ EBA + ∠ BAE

By considering the △ BCD

We know that

x = ∠ CBA + ∠ BCD

It is given that x = y

So we can write it as

∠ CBA + ∠ BCD = ∠ EBA + ∠ BAE

On further calculation we can write it as

∠ BCD = ∠ BAE

Based on both △ BCD and △ BAE

We know that B is the common point

It is given that AB = BC

It is proved that ∠ BCD = ∠ BAE

Therefore, by ASA congruence criterion we get

△ BCD ≅ △ BAE

We know that the corresponding sides of congruent triangles are equal

Therefore, it is proved that AE = CD.

Consider Δ ABC,

We have,

∠B = 70°

And, AB = AC

Therefore,

Δ ABC is an isosceles triangle.

∠B = ∠C (Angle opposite to equal sides are equal)

∠B = ∠C = 70°

And ∠A + ∠B + ∠C = 180°(Angles of triangle)

∠A + 70° + 70° = 180°

∠A = 40°

Sides BA and CA have been produced such that BA = AD and CA = AE. 

To prove: DE ∥ BC

Consider △BAC and △DAE, 

BA = AD and CA= AE (Given) 

∠BAC = ∠DAE (vertically opposite angles)

By SAS congruence criterion, we have 

△BAC ≃ △DAE 

We know, corresponding parts of congruent triangles are equal 

So, BC = DE and ∠DEA = ∠BCA, ∠EDA = ∠CBA 

Now, DE and BC are two lines intersected by a transversal DB s.t. 

∠DEA=∠BCA (alternate angles are equal) 

Therefore, DE ∥ BC. Proved.

In the figure, given that:

RT = TS (i)

∠1 = 2∠2 (ii)

And,

∠4 = 2∠3 (iii)

To prove: ΔRBT ≅ ΔSAT

Let the point of intersection of RB and SA be denoted by O.

Since, RB and SA intersect at O.

∠AOR = ∠BOS (Vertically opposite angle)

∠1 = ∠4

2∠2 = 2∠3 [From (ii) and (iii)]

∠2 = ∠3 (iv)

Now, we have in ΔTRS

RT = TS

ΔTRS is an isosceles triangle

Therefore, ∠TRS = ∠TSI (v)

But, we have

∠TRS = ∠TRB + ∠2 (vi)

∠TSR = ∠TSA + ∠3 (vii)

Putting (vi) and (vii) in (v), we get

∠TRB + ∠2 = ∠TSA + ∠3

∠TRB = ∠TSA [From (iv)]

Now, in ΔRBT and ΔSAT

RT = ST From (i)

∠TRB = ∠TSA From (iv)

∠RTB = ∠STA (Common angle)

By ASA theorem,

ΔRBT ≅ ΔSAT

In △ ABC it is given that ∠ A = 100^o and ∠ C = 50^o

Based on the sum property of the triangle

∠ A + ∠ B + ∠ C = 180^o

To find ∠ B

∠ B = 180^o – ∠ A – ∠ C

By substituting the values in the above equation

∠ B = 180^o – 100^o – 50^o

By subtraction

∠ B = 180^o – 150^o

∠ B = 30^o

So we get ∠ B < ∠ C < ∠ A

i.e. AC < AB < BC

Hence, AC is the shortest side in △ ABC.

Let Δ ABC be isosceles where AB = AC and ∠ B = ∠ C 

Given: Vertex angle A is twice the sum of the base angles B and C. i.e., 

∠ A = 2(∠ B + ∠ C) 

∠ A = 2(∠ B + ∠ B) 

∠ A = 2(2 ∠ B) 

∠ A = 4(∠ B) 

Now, We know that sum of angles in a triangle =180° 

∠ A + ∠ B + ∠ C = 180° 

4 ∠ B + ∠ B + ∠ B = 180°

6 ∠ B =180° 

∠ B = 30° 

Since, ∠ B = ∠ C 

∠ B = ∠ C = 30° 

And ∠ A = 4 ∠ B 

∠ A = 4 x 30° = 120° 

Therefore, angles of the given triangle are 30° and 30° and 120°.

Let the base angles be x each,

Vertex angle = 2(x + x) = 4x

Now, since the sum of all the angles of a triangle is 180°

x + x + 4x = 180°

6x = 180°

x = 30°

Therefore, vertex angle= 4x = 120°

RHS congruence property:- Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.

ΔRPQ ≅ ΔLNM

SAS congruence property:- Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

ΔACB ≅ ΔDEF

ASA congruence property:- Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

ΔACB ≅ ΔACD

SSS congruence property:- Tow triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

ΔYXZ ≅ ΔTRS

ASA congruence property:- Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

ΔDEF ≅ ΔPNM

(a) (i) AR =

AR = PE,

(ii) RT =

RT = EN,

(iii) AT =

AT = PN

(b) 

(i) RT =

RT = EN and

(ii) PN =

PN = AT

(c) 

(i) ∠ATR = ∠PNE and

(ii) ∠TAR = NPE.

Since, two adjacent sides and the so formed angle with the two sides are shown equal while proving them congruent. Hence, by S.A.S. theorem the two triangles can be proved congruent.

In ∆ABC

∠A + ∠B + ∠C = 180°
45° + 65° + ∠C = 180°
110° + ∠C = 180°
∠C = 180° – 110° = 70°

In ∆ABC and ∆PRQ

(i) side PR = side AB = 3.5 cm
(ii) side QR = side CB = 2 cm
(iii) side PQ = side AC = 2.9 cm
(iv) ∠P =∠A = 45°
(v) ∠Q =∠C = 70°
(vi) ∠R = ∠B = 65°

Since, BC and PQ are the non c.p.c.t. part of the two given triangles, hence we cannot judge them to be equal or not from the given information about them.

Since, by corresponding part of congruent triangle ∠E of ΔEFD is equal to the ∠P of ΔPQR.

Since, by corresponding part of congruent triangle ED of ΔEFD is equal to the PR of ΔPQR.

Given that in ΔABC,

∠B = ∠C = 45°

We have to find longest side.

We know that,

Sum of angles in a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A + 45° + 45° = 180°

∠A + 90° = 180°

∠A = 90°

Therefore, BC is the longest side because side opposite to greater angle is larger.