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AB and AC are the equal sides of equilateral triangle ABC because only then after reverting it to form ACB, the two triangles can be proved congruent.

(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.

(ii) We have used Hypotenuse AB = Hypotenuse AC

AD = DA

∠ADB = ∠ADC = 90° (AD ⊥ BC at point D)

(iii) Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)

Since we have already proved that the two triangles are congruent.

Given that in figure,

AD ⊥ CD and CB ⊥ CD

And,

AQ = BP, DP = CQ

We have to prove that,

∠DAQ = ∠CBP

Given that, DP = QC

Adding PQ on both sides, we get

DP + PQ = PQ + QC

DQ = PC (i)

Now consider, ΔDAQ and ΔCBP we have

∠ADQ = ∠BCP = 90°(Given)

AQ = BP (Given)

And,

DQ = PC (From i)

So, by RHS congruence rule, we have

ΔDAQ ≅ ΔCBP

Now,

∠DAQ = ∠CBP (By c.p.c.t)

Hence, proved

Given: PQRS is a square and SRT is an equilateral triangle. 

To prove: (i) PT =QT and (ii) ∠TQR = 15° 

Now, 

PQRS is a square: 

PQ = QR = RS = SP …… (i) 

And ∠ SPQ = ∠ PQR = ∠ QRS = ∠ RSP = 90° 

Also, △ SRT is an equilateral triangle: 

SR = RT = TS …….(ii) 

And ∠ TSR = ∠ SRT = ∠ RTS = 60° 

From (i) and (ii) 

PQ = QR = SP = SR = RT = TS ……(iii) 

From figure, 

∠TSP = ∠TSR + ∠ RSP 

= 60° + 90° 

= 150° and 

∠TRQ = ∠TRS + ∠ SRQ 

= 60° + 90° 

= 150° 

∠ TSR = ∠ TRQ = 150° ………………… (iv) 

By SAS congruence criterion, Δ TSP ≃ Δ TRQ 

We know, corresponding parts of congruent triangles are equal

So, PT = QT

Proved part (i).

Now, consider ΔTQR. 

QR = TR [From (iii)]

Δ TQR is an isosceles triangle. 

∠QTR = ∠TQR [angles opposite to equal sides] 

Sum of angles in a triangle = 180° 

∠QTR + ∠ TQR + ∠TRQ = 180°

2∠TQR + 150° = 180° [From (iv)] 

2 ∠TQR = 30° 

∠TQR = 15°

Hence proved part (ii).

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

In ΔABC,

∠A + ∠B+ ∠C = 180°

∠A = 76°

Now,

∠BAC = ∠ACD (Alternate angles)

∠ACD = 76°

Now,

∠ACD + ∠ECD = 180°

x = 180 - 76°

x = 104°

Given that ΔABC

BE is bisector of ∠Band AD is bisector of ∠BAC

∠B = 2 ∠C

By exterior angle theorem in triangle ADC

∠ADB = ∠DAC + ∠C (i)

In ΔADB,

∠ABD + ∠BAD + ∠ADB = 180°

2 ∠C + ∠BAD + ∠DAC + ∠C = 180°[From (i)]

3 ∠C + ∠BAC = 180°

∠BAC = 180° – 3 ∠C (ii)

Therefore,

AB = CD

∠C = ∠DAC

∠C = \(\frac{1}{2}\) ∠BAC (iii)

Putting value of Angle C in (ii), we get

∠BAC = 180° – \(\frac{1}{2}\) ∠BAC

∠BAC + \(\frac{3}{2}\)∠BAC = 180°

\(\frac{5}{2}\)∠BAC = 180°

∠BAC = \(\frac{180\times 2}{5}\)

= 72°

∠BAC = 72°

∠Z = 90° (Angle of square)

Therefore, AZD is a right angle triangle,

By Pythagoras theorem,

AD² = AZ² + ZD²

AD² = 2² + (2+3)²

AD2 = 4 + 25

AD = √29

In ΔAXB, with x as right angle,

By Pythagoras theorem,

AB² = AX² + XB²

XB² = 29 - 4

XB = 5

BY = XB + XY

= 5 + 2

= 7 cm

Given,

Triangles ABC and DCB are right-angled at A and D respectively.

AC = DB

To prove,

ΔABC = ΔDCB

Proof,

In ΔABC and ΔDCB:

AC = DB (given)

BC = BC (common)

∠CAB = ∠BDC = 90^o

From the RHS congruence property,

ΔABC ≅ ΔDCB.

We know that △ OAB is an equilateral triangle

So it can be written as

∠ OAB = ∠ OBA = AOB = 60^o

From the figure we know that ABCD is a square

So we get

∠ A = ∠ B = ∠ C = ∠ D = 90^o

In order to find the value of ∠ DAO

We can write it as

∠ A = ∠ DAO + ∠ OAB

By substituting the values we get

90^o = ∠ DAO + 60^o

On further calculation

∠ DAO = 90^o – 60^o

By subtraction

∠ DAO = 30^o

We also know that ∠ CBO = 30^o

Considering the △ OAD and △ OBC

We know that the sides of a square are equal

AD = BC

We know that the sides of an equilateral triangle are equal

OA = OB

By SAS congruence criterion

△ OAD ≅ △ OBC

So we get OD = OC (c. p. c. t)

Therefore, it is proved that △ OCD is an isosceles triangle.

Given,

AB = AD and CB = CD

To prove,

ΔABC ≅ ΔADC.

Proof,

In ΔABC and ΔADC

AB = AD (given)

CB = CD (given)

AC = AC (common)

∴ ΔABC ≅ ΔADC.

(By SSS congruence property)

Given,

AB = AC and BD = DC

To prove,

ΔADB ≅ ΔADC

Proof,

In the right triangles ADB and ADC, we have:

Hypotenuse AB = Hypotenuse AC (given)

BD = DC (given)

AD = AD (common)

∴ ΔADB ≅ ΔADC

By SSS congruence property:

∠ADB = ∠ADC (corresponding parts of the congruent triangles) … (1)

∠ADB and ∠ADC are on the straight line.

∴∠ADB + ∠ADC =180^o

∠ADB + ∠ADB = 180^o

2 ∠ADB = 180^o

∠ADB = 180/2

∠ADB = 90^o

From (1):

∠ADB = ∠ADC = 90^o

(ii) ∠BAD = ∠CAD (∵ corresponding parts of the congruent triangles)

In Δ ABC, ∠ A = 40° and ∠ B = 60° 

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

∠ A + ∠ B + ∠ C = 180° 

40° + 60° + ∠ C = 180° 

∠C = 180° – 100° = 80° 

∠ C = 80° 

Now, 40° < 60° < 80° 

∠ A < ∠ B < ∠ C 

∠ C is greater angle and ∠ A is smaller angle. 

Now, ∠ A < ∠ B < ∠ C 

We know, side opposite to greater angle is larger and side opposite to smaller angle is smaller. 

Therefore, BC < AC < AB 

AB is longest and BC is shortest side.

It is given that AB = AC and the bisectors ∠ B and ∠ C meet at a point O

Consider △ BOC

So we get

∠ BOC = ½ ∠ B and ∠ OCB = ½ ∠ C

It is given that AB = AC so we get ∠ B = ∠ C

So we get

∠ OBC = ∠ OCB

We know that if the base angles are equal even the sides are equal

So we get OB = OC ……. (1)

∠ B and ∠ C has the bisectors OB and OC so we get

∠ ABO = ½ ∠ B and ∠ ACO = ½ ∠ C

So we get

∠ ABO = ∠ ACO …….. (2)

Considering △ ABO and △ ACO and equation (1) and (2)

It is given that AB = AC

By SAS congruence criterion

△ ABO ≅ △ ACO

∠ BAO = ∠ CAO (c. p. c. t)

Therefore, it is proved that BO = CO and the ray AO is the bisector of ∠ A.

In Δ ABC, ∠ B = ∠ C = 45° 

Sum of angles in a triangle = 180° 

∠ A + ∠ B + ∠ C = 180° 

∠ A + 45° + 45° = 180° 

∠ A = 180° – (45° + 45°) 

= 180° – 90° 

= 90° 

∠ A = 90° 

∠ B = ∠ C < ∠ A 

Therefore, BC is the longest side.

Given:

Let ABCD is a quadrilateral with AC and BD as its diagonals

To Prove:

Sum of all the sides of a quadrilateral is greater than the sum of its diagonals

Proof:

Consider a quadrilateral ABCD where AC and BD are the diagonals

AB+BC > AC (i) (Sum of two sides is greater than the third side)

AD+DC > AC (ii)

AB+AD > BD (iii)

DC+BC > BD (iv)

Adding (i), (ii), (iii), and (iv)

AB+BC+AD+DC+AB+AD+DC+BC > AC+AC+BD+BD

2(AB+BC+CD+DA) > 2(AC+BD)

AB+BC+CD+DA > AC+BC

Hence, proved that the

Sum of all the sides of a quadrilateral is greater than the sum of its diagonals

(i) In ∆ABC and ∆PQR

AB = PQ= 1.5 cm
BC = QR = 2.5 cm
CA = RP = 2.2 cm
By S.S.S. condition of congruency ∆ABC ≅ ∆PQR

(ii) In ∆DEF and ∆LMN , EF ≠ MN

∴ These triangles are not congruent.

(iii) In ∆ABC and ∆PQR

BC ≠ QR

∴ These triangles are not congruent.

(iv) In ∆ADB and ∆ADC

AD = AD (common)
DB = DC = 2.5 cm
and BA = CA = 3.5 cm

In S.S.S. condition of congruency , ∆ADB ≅ ∆ADC.

Given, Length of sides are 2 cm, 3 cm and 7 cm.

We have to check whether it is possible to draw a triangle with the given length of sides.

We know that, 

A triangle can be drawn only when the sum of any two sides is greater than the third side.

Here,

2 + 3 >7

So, the triangle does not exist.

(i) No. It is not possible to construct a triangle with lengths of its sides 5cm, 4cm and 9cm because the sum of two sides is not greater than the third side i.e. 5 + 4 is not greater than 9.

(ii) Yes. It is possible to construct a triangle with lengths of its sides 8cm, 7cm and 4cm because the sum of two sides of a triangle is greater than the third side.

(iii) Yes. It is possible to construct a triangle with lengths of its sides 10cm, 5cm and 6cm because the sum of two sides of a triangle is greater than the third side.

(iv) Yes. It is possible to construct a triangle with lengths of its sides 2.5cm, 5cm and 7cm because the sum of two sides of a triangle is greater than the third side.

(v) No. It is not possible to construct a triangle with lengths of its sides 3cm, 4cm and 8cm because the sum of two sides is not greater than the third side.

(i) False:

Sum of three sides of a triangle is greater than sum of its three altitudes.

(ii) True

(iii) True

(iv) False:

The difference of any two sides of a triangle is less than the third side.

(v) True:

The side opposite to greater angle is longer and smaller angle is shorter in a triangle. 

(vi) True:

The perpendicular distance is the shortest distance from a point to a line not containing it.

Lengths of sides are 2 cm, 3 cm and 7 cm. 

A triangle can be drawn only when the sum of any two sides is greater than the third side. 

So, let’s check the rule. 

2 + 3 > 7 or 2 + 3 < 7 

2 + 7 > 3 and 3 + 7 > 2 

Here 2 + 3 >7 

So, the triangle does not exit.

By c.p.c.t. that is corresponding part of congruent triangles, the pair of equal angles are:

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

(i) They are of equal lengths

(ii) Their measures are the same or equal.

(iii) Their sides are equal or they have the same side length

(iv) Their dimensions are same that is lengths are equal and their breadths are also equal.

(v) They have same radii

In ΔABD and ΔFEC,

AB = FE (Given)

∠B = ∠E (Each 90°)

BC = DE (Given)

Add CD both sides, we get

BD = EC

Therefore, by S.A.S. theorem,

ΔABD ≅ ΔFEC

(a) They have the same measure, (length)

(b) 70°

(c) m < A ≅ m

Given,

AB = AC

∠ACD = 105°

Since,

∠BCD = 180°(Straight angle)

∠BCA + ∠ACD = 180°

∠BCA + 105° = 180°

∠BCA = 75°(i)

Now,

ΔABC is an isosceles triangle

∠ABC = ∠ACB (Angle opposite to equal sides)

From (i), we have

∠ACB = 75°

∠ABC = ∠ACB = 75°

Sum of interior angle of triangle = 180°

∠A +∠B +∠C =180°

∠A = 180° - 75° -75°

= 30°

Therefore,

∠BAC = 30°

Consider the figure,

Given,

AB = AC

DB = DC

To find: Ratio

∠ABD =∠ACD

Now, ΔABC and ΔDBC are isosceles triangles

Since,

AB = AC

And,

DB = DC

Therefore,

∠ABC = ∠ACB and,

∠DBC = ∠DCB (Angle opposite equal sides)

Now, consider ∠ABD : ∠ACD

(∠ABC - ∠DBC) : (∠ACB - ∠DCB)

(∠ABC - ∠DBC) : (∠ABC - ∠DBC) [Since, ∠ABC = ∠ACB and ∠DBC = ∠DCB]

1: 1

Therefore,

∠ABD : ∠ACD = 1:1