Congruence Of Triangles And Inequalities In A Triangle

Given that

AB = AC and ∠A = 70°

To find: ∠B and ∠C

AB = AC and also ∠A = 70°

As two sides of triangle are equal, we say that ∆ABC is isosceles triangle.

Hence by the property of isosceles triangle, we know that base angles are also equal.

ie. we state that ∠B = ∠C. …(1)

Now,

Sum of all angles in any triangle = 180°

∴ ∠A + ∠B + ∠C = 180°

Hence,

70° + ∠B + ∠C = 180°

2 ∠B = 180° - 70° …from (1)

∴ 2∠B= 110°

∠B = 55°

Therefore, our base angles, ∠B and ∠C, are 55° each.

Given: The given triangle is isosceles triangle. Also vertex angle is 100°

To find: Measure of base angles.

It is given that triangle is isosceles.

So let our given triangle be ∆ABC.

And let ∠A be the vertex angle, which is given as ∠A= 100°

By the property of isosceles triangle, we know that base angles are equal.

So,

∠B = ∠C …(1)

We know that,

Sum of all angles in any triangle = 180°

∴ ∠A + ∠B + ∠C = 180°

100° + 2∠B = 180° …from (1)

∴ 2∠B = 180° - 100°

2∠B = 80°

∴∠B = 40°

Therefore, our base angles, ∠B and ∠C, are 40° each.

Given: In ∆ABC,

AB=AC and ∠B=65°

To find : ∠A and ∠C

It is given that AB=AC and ∠B=65°

As two sides of the triangle are equal, we say that triangle is isosceles triangle, with vertex angle A.

Hence by the property of isosceles triangle we know that base angles are equal.

∴ ∠B = ∠C

∴∠C = ∠B =65°

Also, We know that,

Sum of all angles in any triangle = 180°

∴∠A + ∠B + ∠C = 180°

∠A + 65° + 65° = 180°

∠A + 130° = 180°

∴∠A = 180° - 130°

∠A = 50°

Hence, ∠C = 65° and ∠A = 50°

Given: Our given triangle is isosceles triangle. Also, the vertex angle is twice the sum of the base angles

To find: Measures of angles of triangle.

It is given that that given triangle is isosceles triangle.

Let vertex angle be y and base angles be x each.

So by given condition,

y = 2(x + x)

∴ y = 4x

Also, We know that,

Sum of all angles in any triangle = 180°

∴ y + x + x = 180°

y +2x = 180°

4x + 2x = 180°

∴ 6x = 180°

x = 30°

∴ y = 4 × 30°

y = 120°

Hence, vertex angle is 120° and base angles are 30° each.

Here given triangle is isosceles right angled triangle.

So let our triangle be ∆ABC, right angled at A.

∴ ∠A = 90°

Here, AB = AC, as our given triangle is isosceles triangle.

Hence, base angles, ∠B and ∠C are equal.

Also, We know that,

Sum of all angles in any triangle = 180°

∴ ∠A + ∠B + ∠C = 180°

90° + 2 ∠B = 180°

2∠B = 90°

∠B = 45°

Hence the measure of each of the equal angles of a right-angled isosceles triangle is 45°

Given: ∆ABC is isosceles triangle.

To prove: ∠CAD = ∠CBE

Let ∆ABC be our isosceles triangle as shown in the figure.

We know that base angles of the isosceles triangle are equal.

Here, ∠CAB = ∠CBA ….(1)

Also here, ∠CAD and ∠CBE are exterior angles of the triangle.

So, we know that,

∠CAB +∠CAD = 180°… exterior angle theorem

And ∠CBA + ∠CBE = 180° … exterior angle theorem

So from (1) and above statement, we conclude that,

∠CAB +∠CAD = 180°

And ∠CAB +∠CBE = 180°

Which implies that,

∠CAD = 180° - ∠CAB

And ∠CBE = 180° - ∠CAB

Hence we say that ∠CAD = ∠CBE

∴For the isosceles triangle, the exterior angles so formed are equal to each other.

Given: ∆ABC is equilateral triangle.

To prove: ∠CAD = ∠CBE = ∠BCL

Proof:

Let our triangle be ∆ABC, which is equilateral triangle as shown in the figure.

Hence all angles are equal and measure 60° each.

∴ ∠CAB = ∠CBA = ∠BCA = 60° …(1)

Also here, ∠CAD and ∠CBE are exterior angles of the triangle.

So, we know that,

∠CAB +∠CAD = 180° … exterior angle theorem

∠CBA + ∠CBE = 180° … exterior angle theorem

∠BCA + ∠BCL = 180° … exterior angle theorem

From (1) and above statements, we state that,

60° +∠CAD = 180°

60° + ∠CBE = 180°

60° + ∠BCL = 180°

Simplifying above statements,

∠CAD = 180° - 60° = 120°

∠CBE = 180° - 60° = 120°

∠BCL = 180° - 60° = 120°

Hence, the measure of each exterior angle of an equilateral triangle is 120°

Given: AO = OB , DO = OC

To prove: AC=BD and AC||BD

Proof:

It is given that, O is the midpoint of each of the line segments AB and CD.

This implies that AO = OB and DO = OC

Here line segments AB and CD are concurrent.

So,

∠AOC = ∠BOD …. As they are vertically opposite angles.

Now in ∆AOC and ∆BOD,

AO = OB,

OC = OD

Also, ∠AOC = ∠BOD

Hence, ∆AOC ≅ ∆BOD … by SAS property of congruency

So,

AC = BD … by cpct

∴ ∠ACO = ∠BDO … by cpct

But ∠ACO and ∠BDO are alternate angles.

∴ We conclude that AC is parallel to BD.

Hence we proved that AC=BD and AC||BD

Given: PA ⊥ AB, QB ⊥ AB and PA=QB

To prove: AO = OB and PO = OQ

It is given that PA ⊥ AB and QB ⊥ AB.

This means that ∆PAO and ∆QBO are right angled triangles.

It is also given that PA=QB

Now in ∆PAO and ∆QBO,

∠OAP = ∠OBQ = 90°

PO = OQ

Hence by hypotenuse-leg congruency,

∆PAO ≅ ∆QBO

∴AO = OB and PO = OQ ….by cpct

Hence proved that AO = OB and PO = OQ

Given: AO = OD and CO = OB

To prove: AC = BD

Proof :

It is given that AO = OD and CO = OB

Here line segments AB and CD are concurrent.

So,

∠AOC = ∠BOD …. As they are vertically opposite angles.

Now in ∆AOC and ∆DOB,

AO = OD,

CO = OD

Also, ∠AOC = ∠BOD

Hence, ∆AOC ≅ ∆BOD … by SAS property of congruency

So,

AC = BD … by cpct

Here,

∠ACO ≠ ∠BDO or ∠OAC ≠ ∠OBD

Hence there are no alternate angles, unless both triangles are isosceles triangle.

Hence proved that AC=BD but AC may not be parallel to BD.

Here it is given that l ‖ m ie. AC ||DB.

Also given that AM = MB

Now in ∆AMC and ∆BMD,

∠CAM = ∠DBM … Alternate angles

AM = MB

∠AMC = ∠BMD … vertically opposite angles

Hence, ∆AMC ≅ ∆BMD … by ASA property of congruency

∴ CM = MD …cpct

Hence proved that M is also the midpoint of any line segment CD having its end points atﺎ and m respectively.

∆ABC and ∆OBC are isosceles triangle.

∴ ∠ABC = ∠ACB and ∠OBC = ∠OCB ….(1)

Also,

∠ABC = ∠ABO + ∠OBC

And ∠ACB = ∠ACO + ∠OCB

From 1 and above equations, we state that,

∠ABC = ∠ABO + ∠OBC

And ∠ABC = ∠ACO + ∠OBC

This implies that,

∠ABO = ∠ABC - ∠OBC

And ∠ACO = ∠ABC - ∠OBC

Hence,

∠ABO = ∠ACO = ∠ABC - ∠OBC

Given that AB = AC and also DE || BC.

So by Basic proportionality theorem or Thales theorem,

=

∴ =

Now adding 1 on both sides,

+ 1 = + 1

=

= … as AB = AD + DE and AC = AE + EC

But is given that AB = AC,

∴ =

Hence,

AD = AE.

Here it is given that AX = AY.

Now in ∆CXA and ∆BYA,

AX = AY

∠XAC = ∠YAB … Same angle or common angle.

AC = AB … given condition Hence by SAS property of congruency,

∆CXA ≅ ∆BYA

Hence by cpct, we conclude that,

CX = BY

It is given that AC = BC , ∠DCA = ∠ECB and ∠DBC = ∠EAC.

Adding angle ∠ECD both sides in ∠DCA = ∠ECB, we get,

∠DCA + ∠ECD = ∠ECB + ∠ECD

∴∠ECA = ∠DCB …addition property

Now in ΔDBC and ΔEAC,

∠ECA = ∠DCB

BC = AC

∠DBC = ∠EAC

Hence by ASA postulate, we conclude,

ΔDBC ≅ ΔEAC

Hence, by cpct, we get,

DC = EC

Given : BA ⊥ AC and DE ⊥ EF such that BA=DE and BF=DC

To prove: AC = EF

Proof:

In ∆ABC, we have,

BC = BF + FC

And , in ∆DEF,

FD = FC + CD

But, BF = CD

So, BC = BF + FC

And, FD = FC +BF

∴ BC = FD

So, in ∆ABC and ∆DEF, we have,

∠BAC = ∠DEF … given

BC = FD

AB = DE …given

Thus by Right angle - Hypotenuse- Side property of congruence, we have,

∆ABC ≅ ∆DEF

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ AC = EF

Given: x=y and AB=CB

To prove: AE = CD

Proof:

In ∆ABE, we have,

∠AEC = ∠EBA + ∠BAE …Exterior angle theorem

y° = ∠EBA + ∠BAE

Now in ∆BCD, we have,

x° = ∠CBA + ∠BCD

Since, given that,

x = y ,

∠CBA + ∠BCD = ∠EBA + ∠BAE

∴ ∠BCD = ∠BAE … as ∠CBA and ∠EBA and same angles.

Hence in ∆BCD and ∆BAE,

∠B = ∠B

BC = AB …given

∠BCD = ∠BAE

Thus by ASA property of congruence, we have,

∆BCD ≅ ∆BAE

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ CD = AE

Given: AB=AC and BD and AB are angle bisectors of ∠B and ∠C

To prove: BD = CE

Proof:

In ∆ABD and ∆ACE,

∠ABD = ∠B

And ∠ACE = ∠C

But ∠B = ∠C as AB = AC … As in isosceles triangle, base angles are equal

∠ABD = ∠ACE

AB = AC

∠A = ∠A

Thus by ASA property of congruence,

∆ABD ≅ ∆ACE

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ BD = CE

Given: BC = DC and BL ⊥ AD and DM ⊥ CM

To prove: BL=CM

Proof:

In ∆BLD and ∆CMD,

∠BLD = ∠CMD = 90° … given

∠BLD = ∠MDC … vertically opposite angles

BD = DC … given

Thus by AAS property of congruence,

∆BLD ≅ ∆CMD

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ BL = CM

Given: BD = DC and DL ⊥ AB and DM ⊥ AC such that DL=DM

To prove: AB = AC

Proof:

In right angled triangles ∆BLD and ∆CMD,

∠BLD = ∠CMD = 90°

BD = CD … given

DL = DM … given

Thus by right angled hypotenuse side property of congruence,

∆BLD ≅ ∆CMD

Hence, we know that, corresponding parts of the congruent triangles are equal

∠ABD = ∠ACD

In ∆ABC, we have,

∠ABD = ∠ACD

∴ AB= AC …. Sides opposite to equal angles are equal

Given: In ∆ABC, AB=AC and the bisectors of ∠B and ∠C meet at a point O.

To prove: BO=CO and ∠BAO = ∠CAO

Proof:

In , ∆ABC we have,

∠OBC = ∠B

∠OCB = ∠C

But ∠B = ∠C … given

So, ∠OBC = ∠OCB

Since the base angles are equal, sides are equal

∴ OC = OB …(1)

Since OB and OC are bisectors of angles ∠B and ∠C respectively, we have

∠ABO = ∠B

∠ACO = ∠C

∴∠ABO = ∠ACO …(2)

Now in ∆ABO and ∆ACO

AB = AC … given

∠ABO = ∠ACO … from 2

BO = OC … from 1

Thus by SAS property of congruence,

∆ABO ≅ ∆ACO

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BAO = ∠CAO

ie. AO bisects ∠A

Given: PQR is an equilateral triangle and QRST is a square

To prove: PT=PS and ∠PSR=15°.

Proof:

Since ∆PQR is equilateral triangle,

∠PQR = ∠PRQ = 60°

Since QRTS is a square,

∠RQT = ∠QRS = 90°

In ∆PQT,

∠PQT = ∠PQR + ∠RQT

= 60° + 90°

= 150°

In ∆PRS,

∠PRS = ∠PRQ + ∠QRS

= 60° + 90°

= 150°

∴ ∠PQT = ∠PRS

Thus in ∆PQT and ∆PRS,

PQ = PR … sides of equilateral triangle

∠PQT = ∠PRS

QT = RS … side of square

Thus by SAS property of congruence,

∆PQT ≅ ∆PRS

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PT = PS

Now in ∆PRS, we have,

PR = RS

∴ ∠PRS = ∠PSR

But ∠PRS = 150°

SO, by angle sum property,

∠PRS + ∠PSR + ∠SPR = 180°

150° + ∠PSR + ∠SPR = 180°

2∠PSR = 180° - 150°

2∠PSR = 30°

∠PSR = 15°

Given: ∠ABC = 90° , BCDE is a square on side BC and ACFG is a square on AC

To prove: AD = EF

Proof:

Since BCDE is square,

∠BCD = 90° …(1)

In ∆ACD,

∠ACD = ∠ACB + ∠BCD

= ∠ACB + 90° …(2)

In ∆BCF,

∠BCF = ∠BCA + ∠ACF

Since ACFG is square,

∠ACF = 90° …(3)

From 2 and 3, we have,

∠ACD = ∠BCF ….(4)

Thus in ∆ACD and ∆BCF, we have,

AC = CF ...sides of square

∠ACD = ∠BCF …from 4

CD = BC … sides of square

Thus by SAS property of congruence,

∆ACD ≅ ∆BCF

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ AD = BF

Given: ∆ABC is isosceles triangle where AB = AC and BD = DC

To prove: ∠BAD = ∠DAC

Proof:

In ∆ABD and ∆ADC

AB = AC …given

BD = DC …given

AD = AD … common side

Thus by SSS property of congruence,

∆ABD ≅ ∆ADC

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BAD = ∠DAC

Given: ABCD is a quadrilateral in which AB‖DC and BP = PC

To prove: AB=CQ and DQ=DC+AB

Proof:

In ∆ABP and ∆PCQ we have,

∠PAB = ∠PQC …alternate angles

∠APB = ∠CPQ … vertically opposite angles

BP = PC … given

Thus by AAS property of congruence,

∆ABP ≅ ∆PCQ

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ AB = CQ …(1)

But, DQ = DC + CQ

= DC + AB …from 1

Given: OA=OB and OP=OQ

To prove: PX=QX and AX=BX

Proof:

In ∆OAQ and ∆OPB, we have

OA = OB …given

∠O = ∠O …common angle

OQ = OP … given

Thus by SAS property of congruence,

∆OAP ≅ ∆OPB

Hence, we know that, corresponding parts of the congruent triangles are equal

∠OBP = ∠OAQ …(1)

Thus, in ∆BXQ and ∆PXA, we have,

BQ = OB – OQ

And PA = OA – OP

But OP = OQ

And OA = OB …given

Hence, we have, BQ = PA …(2)

Now consider ∆BXQ and ∆PXA,

∠BXQ = ∠PXA … vertically opposite angles

∠OBP = ∠OAQ …from 1

BQ = PA … from 2

Thus by AAS property of congruence,

∆BXQ ≅ ∆PXA

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PX = QX

And AX = BX

Given: ABCD is a square and PB=PD

To prove: CPA is a straight line

Proof:

∆APD and ∆APB,

DA = AB …as ABCD is square

AP = AP … common side

PB = PD … given

Thus by SSS property of congruence,

∆APD ≅ ∆APB

Hence, we know that, corresponding parts of the congruent triangles are equal

∠APD = ∠APB …(1)

Now consider ∆CPD and ∆CPB,

CD = CB … ABCD is square

CP = CP … common side

PB = PD … given

Thus by SSS property of congruence,

∆CPD ≅ ∆CPB

Hence, we know that, corresponding parts of the congruent triangles are equal

∠CPD = ∠CPB … (2)

Now,

Adding both sides of 1 and 2,

∠CPD +∠APD = ∠APB + ∠CPB …(3)

Angels around the point P add upto 360°

∴ ∠CPD +∠APD + ∠APB + ∠CPB = 360°

From 4,

2(∠CPD +∠APD) = 360°

∠CPD +∠APD = = 180°

This proves that CPA is a straight line.

Given: ABC is an equilateral triangle, PQ ‖AC and CR=BP

To prove: QR bisects PC or PM = MC

Proof:

Since, ∆ABC is equilateral triangle,

∠A = ∠ACB = 60°

Since, PQ ‖AC and corresponding angles are equal,

∠BPQ = ∠ACB = 60°

In ∆BPQ,

∠B= ∠ACB = 60°

∠BPQ = 60°

Hence, ∆BPQ is an equilateral triangle.

∴ PQ = BP = BQ

Since we have BP = CR,

We say that PQ = CR …(1)

Consider the triangles ∆PMQ and ∆CMR,

∠PQM = ∠CRM …alternate angles

∠PMQ = ∠CMR … vertically opposite angles

PQ = CR … from 1

Thus by AAS property of congruence,

∆PMQ ≅ ∆CMR

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PM = MC

Given: ABCD is a quadrilateral in which AB=AD and BC=DC

To prove: AC bisects ∠A and ∠C, and AC is the perpendicular bisector of BD

Proof:

In ∆ABC and ∆ADC, we have

AB = AD …given

BC = DC … given

AC = AC … common side

Thus by SSS property of congruence,

∆ABC ≅ ∆ADC

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BAC = ∠DAC

∴ ∠BAO = ∠DAO …(1)

It means that AC bisects ∠BAD ie ∠A

Also, ∠BCA = ∠DCA … cpct

It means that AC bisects ∠BCD, ie ∠C

Now in ∆ABO and ∆ADO

AB = AD …given

∠BAO = ∠DAO … from 1

AO = AO … common side

Thus by SAS property of congruence,

∆ABO ≅ ∆ADO

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BOA = ∠DAO

But ∠BOA + ∠DAO = 180°

2∠BOA = 180°

∴ ∠BOA = = 90°

Also ∆ABO ≅ ∆ADO

So, BO = OD

Which means that AC = BD

Given: IP ⊥BC, IQ ⊥CA and IR ⊥ AB and the bisectors of ∠B and ∠C of ∆ABC meet at I

To prove: IP=IQ=IR and IA bisects ∠A

Proof:

In ∆BIP and ∆BIR we have,

∠PBI = ∠RBI …given

∠IRB = ∠IPB = 90° …Given

IB = IB …common side

Thus by AAS property of congruence,

∆BIP ≅ ∆BIR

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ IP = IR

Similarly,

IP = IQ

Hence, IP = IQ = IR

Now in ∆AIR and ∆AIQ

IR = IQ …proved above

IA = IA … Common side

∠IRA = ∠IQA = 90°

Thus by SAS property of congruence,

∆AIR ≅ ∆AIQ

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ ∠IAR = ∠IAQ

This means that IA bisects ∠A

Given: P is a point in the interior of ∠AOB and PL ⊥ OA and PM ⊥ OB such that PL=PM

To prove: ∠POL = ∠POM

Proof:

In ∆OPL and ∆OPM, we have

∠OPM = ∠OPL = 90° …given

OP = OP …common side

PL = PM … given

Thus by Right angle hypotenuse side property of congruence,

∆OPL ≅ ∆OPM

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ ∠POL = ∠POM

Ie. OP is the bisector of ∠AOB

Given: ABCD is a square, AM = MB and PQ ⊥ CM

To prove: PA=BQ and CP=AB+PA

Proof:

In ∆AMP and ∆BMQ, we have

∠AMP = BMQ …vertically opposite angle

∠PAM = ∠MBQ = 90° …as ABCD is square

AM = MB …given

Thus by AAS property of congruence,

∆AMP ≅ ∆BMQ

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PA = BQ and MP = MQ …(1)

Now in ∆PCM and ∆QCM

PM = QM … from 1

∠PMC = ∠QMC … given

CM = CM … common side

Thus by AAS property of congruence,

∆PCM ≅ ∆QCM

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ PC = QC

PC = QB + CB

PC = AB + PA …as AB = CB and PA = QB

Given: AB ⊥ BO and NM ⊥ OM

In ∆ABO and ∆NMO,

∠OBA = ∠OMN

OB = OM …O is mid point of BM

∠BOA = ∠MON …vertically opposite angles

Thus by AAS property of congruence,

∆ABO ≅ ∆NMO

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ AB = MN

Hence, we can calculate the width of the river by calculating MN

Given: ∠A=36° and ∠B=64°

To find: The longest and shortest sides of the triangle

Given that ∠A=36° and ∠B=64°

Hence, by the angle sum property in ∆ABC, we have

∠A + ∠B + ∠C = 180°

36° + 64°+ ∠C = 180°

100° + ∠C = 180°

∠C = 80°

So, we have ∠A=36°, ∠B=64° and ∠C = 80°

∴∠C is largest and ∠A is shortest

Hence,

Side opposite to ∠C is longest.

∴AB is longest

Side opposite to ∠A is shortest.

∴ BC is shortest

It is given that ∠A=90°.

In right angled triangle at 90°

Sum of all angles in triangle is 180°, so other two angles must be less that 90°

So, other angles are smaller than ∠A.

Hence ∠A is largest angle.

We know that side opposite to largest angle is largest.

∴BC is longest side, which is opposite to ∠A.

In ∆ABC given that ∠A=∠B=45°

So, by the angle sum property in ∆ABC, we have

∠A + ∠B + ∠C = 180°

45° + 45° + ∠C = 180°

90° + ∠C = 180°

∴ ∠C = 180° - 90°

∠C = 90°

Hence, largest angle is ∠C

We know that side opposite to largest angle is longest, which is AB

Hence our longest side is AB

Given: In ∆ABC, BD=BC and ∠B=60° and ∠A=70°

To prove: AD>CD and AD>AC

Proof:

In ∆ABC, by the angle sum property, we have

∠A + ∠B + ∠C = 180°

70° + 60° + ∠C = 180°

130° + ∠C = 180°

∠C = 50°

Now in ∆BCD we have,

∠CBD = ∠DAC + ∠ACB … as ∠CBD is the exterior angle of ∠ABC

= 70° + 50°

Since BC = BD …given

So, ∠BCD = ∠BDC

∴ ∠BCD + ∠BDC = 180° - ∠CBD

= 180° - 120° = 60°

2∠BCD = 60°

∠BCD = ∠BDC = 30°

Now in ∆ACD we have

∠A = 70°, ∠D = 30°

And ∠ACD = ∠ACB + ∠BCD

= 50° + 30° = 80°

∴∠ACD is greatest angle

So, the side opposite to largest angle is longest, ie AD is longest side.

∴AD > CD

Since, ∠BDC is smallest angle,

The side opposite to ∠BDC, ie AC, is the shortest side in ∆ACD.

∴AD > AC

Given: In ∆ABC, ∠B=35°,∠C=65° and ∠BAX = ∠XAC

To find: Relation between AX, BX and CX in descending order.

In ∆ABC, by the angle sum property, we have

∠A + ∠B + ∠C = 180°

∠A + 35° + 65° = 180°

∠A + 100° = 180°

∴ ∠A = 80°

But ∠BAX = ∠A

= × 80° = 40°

Now in ∆ABX,

∠B = 35°

∠BAX = 40

And ∠BXA = 180° - 35° - 40°

= 105°

So, in ∆ABX,

∠B is smallest, so the side opposite is smallest, ie AX is smallest side.

∴ AX < BX …(1)

Now consider ∆AXC,

∠CAX = × ∠A

=× 80° = 40°

∠AXC = 180° - 40° - 65°

= 180° - 105° = 75°

Hence, in ∆AXC we have,

∠CAX = 40°, ∠C = 65°, ∠AXC =75°

∴∠CAX is smallest in ∆AXC

So the side opposite to ∠CAX is shortest

Ie CX is shortest

∴ CX <AX …. (2)

From 1 and 2 ,

BX > AX > CX

This is required descending order

Given: ∠BAD = ∠DAC

To prove: AB>BD and AC>DC

Proof:

In ∆ACD,

∠ADB = ∠DAC + ∠ACD … exterior angle theorem

= ∠BAD + ∠ACD … given that ∠BAD = ∠DAC

∠ADB > ∠BAD

The side opposite to angle ∠ADB is the longest side in ∆ADB

So, AB > BD

Similarly in ∆ABD

∠ADC = ∠ABD + ∠BAD … exterior angle theorem

= ∠ABD + ∠CAD … given that ∠BAD = ∠DAC

∠ADC > ∠CAD

The side opposite to angle ∠ADC is the longest side in ∆ACD

So, AC > DC

Given: AB=AC

To prove: AD>AC

Proof:

In ∆ABC,

∠ACD = ∠B + ∠BAC

= ∠ACB + ∠BAC …as ∠C = ∠B as AB = AC

= ∠CAD + ∠CDA +∠BAC …as ∠ACB = ∠CAD + ∠CDA

∴∠ACD > ∠CDA

So the side opposite to ∠ACD is the longest

∴AD > AC

Given: AC>AB and ∠BAD = ∠DAC

To prove: ∠ADC>∠ADB

Proof:

Since AC > AB

∠ABC > ∠ACB

Adding ∠A on both sides

∠ABC + ∠A > ∠ACB + ∠A

∠ABC + ∠BAD > ∠ACB + ∠DAC … As AD is a bisector of ∠A

∴ ∠ADC > ∠ADB

Given: S is any point on the side QR

To prove: PQ+QR+RP>2PS.

Proof:

Since in a triangle, sum of any two sides is always greater than the third side.

So in ∆PQS, we have,

PQ + QS > PS …(1)

Similarly, ∆PSR, we have,

PR + SR > PS …(2)

Adding 1 and 2

PQ + QS + PR + SR > 2PS

PQ + PR + QR > 2PS …as PR = QS +SR

Given: XOY is a diameter and XZ is any chord of the circle.

To prove: XY>XZ

Proof:

In ∆XOZ,

OX + OZ > XZ … sum of any sides in a triangle is a greater than its third side

∴ OX + OY > XZ … As OZ = OY, radius of circle

Hence, XY > XZ …As OX + OY = XY

Given: O is a point within ∆ABC

To prove:

(i) AB+AC>OB+OC

(ii) AB+BC+CA>OA+OB+OC

(iii) OA+OB+OC>(AB+BC+CA)

Proof:

In ∆ABC,

AB +AC >BC ….(1)

And in ∆OBC,

OB + OC > BC …(2)

Subtracting 1 from 2 we get,

(AB + AC) – (OB + OC ) > (BC – BC )

Ie AB + AC > OB + OC

From ׀, AB + AC > OB + OC

Similarly, AB + BC > OA + OC

And AC + BC > OA + OB

Adding both sides of these three inequalities, we get,

(AB + AC ) + (AB + BC) + (AC + BC) > (OB + OC) + (OA + OC) + (OA + OB)

Ie. 2(AB + BC + AC ) > 2(OA + OB + OC)

∴ AB + BC + OA > OA + OB + OC

In ∆OAB,

OA + OB > AB …(1)

In ∆OBC,

OB + OC > BC …(2)

In ∆OCA

OC + OA > CA …(3)

Adding 1,2 and 3,

(OA + OB) + (OB + OC) + (OC+ OA) >AB + BC +CA

Ie. 2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OC > ( AB + BC + CA)

Our given lengths are AB=3cm, BC=3.5cm and CA=6.5cm.

∴ AB + BC = 3 + 3.5 = 6.5 cm

But CA = 6.5 cm

So, AB + BC = CA

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

So, a triangle cannot be drawn with such lengths

From the above given four options, SSA is not a criterion for the congruence of triangles

∴ Option (A) is correct

It is given in the question that,

AB = QR

BC = RP

And, CA = PQ

∴ By SSS congruence criterion

Hence, option (B) is correct

According to the condition given in the question,

If and is not congruent to

Then, clearly BC PQ

∴ It is false

Hence, option (A) is correct

It is given in the question that,

where,

AB = 5 cm

FD = 5 cm

∠ B = 40°

∠ A = 80°

We know that sum of all angles of a triangle is equal to 180⁰

∴ ∠ A + ∠ B + ∠ C = 180^o

80^o + 40^o + ∠ C = 180^o

∠ C = 180^o - 120⁰

= 60^o

As, Angle C = Angle E

∴ Angle E = 60^o

Hence, option (B) is correct

It is given in the question that,

In

AB = 2.5 cm

BC = 6 cm

We know that, the length of a side must be less than the sum of the other two sides

Let us assume the side of AC be x cm

∴ x < 2.5 + 6

x < 8.5

Also, we know that the length of a side must be greater then the difference between the other two sides

∴ x > 6 – 2.5

x > 3.5

Hence, the limits of the value of x is

3.5 < x < 8.5

∴ It is clear the length of AC cannot be 3.4 cm

Hence, option (A) is correct

It is given in the question that,

In , ∠A = 40^o

∠B = 60^o

We know that, sum of all angles of a triangle is equal to 180⁰

∴ ∠A + ∠B + ∠C = 180^o

60^o + 40^o + ∠C = 180^o

∠C = 180^o – 100^o

∠C = 80^o

Hence, the side which is opposite to ∠C is the longest side of the triangle

∴ Option (C) is correct

It is given in the question that,

In , we have

∠B = 35^o

∠C = 65^o

Also the bisector AD of ∠BAC meets at D

∴ ∠A + ∠B + ∠C = 180^o

∠A + 35^o + 65^o = 180^o

∠A = 180^o – 100^o

∠A = 80^o

As, AD is the bisector of ∠BAC

∴ ∠BAD = ∠CAD = 40^o

In , we have

∠BAD > ∠ABD

BD > AD

Also, in

∠ACD > ∠CAD

AD > CD

Hence, BD > AD > CD

∴ Option (B) is correct

From the given figure, we have

AB > AC

∴ ∠ACB > ∠ABC

Also, ∠ADB > ACD

∠ADB > ACB > ∠ABC

∠ADB > ∠ABD

∴ AB > AD

Hence, option (C) is correct

From the given figure, we have

AB > AC

Also, ∠C > ∠B

∠OCB > ∠OBC (Given)

∴ OB > OC

Hence, option (C) is correct

It is given in the question that,

In ∆OAB and ∆OAC, we have

AB = AC

OB = OC

OA = OA (Common)

∴ By SSS congruence criterion

∴ ∠ABO = ∠ACO

So, ∠ABO: ∠ACO = 1: 1

Hence, option (A) is correct

It is given in the question that,

In , we have

∠C > ∠B

We know that, side opposite to the greater angle is larger

∴ AB >AC

Hence, option (B) is correct

From the given question, we have

In ∆OAB, ∆OBC and ∆OCA we have:

OA + OB > AB

OB + OC > BC

And, OC + OA > AC

Adding all these, we get:

2 (OA + OB + OC) > (AB + BC + CA)

(OA + OB + OC >

∴ Option (C) is correct

It is given in the question that,

In ∆ABC, BL is parallel to AC

Also, CM is parallel AB such that BL = CM

We have to prove that: AB = AC

Now, in ∆ABL and ∆ACM we have:

BL = CM (Given)

∠BAL = ∠CAM (Common)

∠ALB = ∠AMC (Each angle equal to 90^o)

∴ By AAS congruence criterion

∆ABL ≅ ∆ACM

AB = AC (By Congruent parts of congruent triangles)

As opposite sides of the triangle are equal, so it is an isosceles triangle

Hence, option (B) is correct

From the given figure, we have

AE = DB

And, CB = EF

Now, AB = (AD – DB)

= (AD – AE)

DE = (AD – AE)

Now, in ∆ABC and ∆DEF we have:

AB = DE

CB = EF

∠ABC = ∠FED

∴ By SAS congruence criterion

∆ABC ≅ ∆DEF

Hence, option (A) is correct

From the given figure, we have

BE is perpendicular to CA

Also, CF is perpendicular to BA

And, BE = CF

Now, in ∆ABE and ∆ACF we have:

BE = CF (Given)

∠BEA = ∠CFA = 90^o

∠A = ∠A (Common)

∴ By AAS congruence criterion

∆ABE ≅ ∆ACF

Hence, option (A) is correct

From the given figure, we have

D is the mid-point of BC

Also, DE is perpendicular to AB

DF is perpendicular to AC

And, DE = DF

Now, in ∆BED and ∆CFD we have:

DE = DF

BD = CD

∠E = ∠F = 90^o

∴ By RHS congruence rule

∆BED ≅ ∆CFD

Thus, ∠B = ∠C

AC = AB

Hence, option (A) is correct

From the question, we have:

In ∆ABC and ∆DEF

AB = DE (Given)

BC = EF (Given)

So, in order to have ∆ABC ≅ ∆DEF

∠B must be equal to ∠E

∴ Option (B) is correct

From the question, we have:

In ∆ABC and ∆DEF

∠B = ∠E (Given)

∠C = ∠F (Given)

So, in order to have ∆ABC ≅ ∆DEF

BE must be equal to EF

∴ Option (C) is correct

It is given in the question that,

In ∆ABC and ∆PQR, we have

AB = AC

Also, ∠C = ∠B

As, ∠C = ∠P and, ∠B = ∠Q

∴ ∠P = ∠Q

So, both triangles are isosceles but not congruent

Hence, option (A) is correct

We know that,

Sum of all angles of a triangle is equal to 180^o

∴ A triangle can have two acute angles because sum of two acute angles of a triangle is always less than 180^o

Thus, it satisfies the angle sum property of a triangle

Hence, option (C) is correct

Here we can clearly see that the true statements are as follows:

(I) In a ∆ABC in which AB=AC, the altitude AD bisects BC.

(II) If the altitudes AD, BE and CF of ∆ABC are equal, then ∆ABC is equilateral.

∴ Option C is correct

According to the question,

In ΔABD and ΔACD,

Since, sum of any two sides of a triangle is greater than the third side.

AB + DB > AD (i)

AC + DC > AD (ii)

Adding (i) and (ii)

AB + AC + DB + DC > 2AD

AB + AC + BC > 2AD

Hence, the assertion and the reason are both true, but Reason does not explain the assertion.

∴ Option B is correct

Since, sum of two sides is greater than the third side

∴AB + BC > AC (i)

CD + DA > AC (ii)

Adding (i) and (ii),

AB + BC +CD +DA > 2AC

Hence, the assertion is true and also the reason gives the right explanation of the assertion.

∴ Option A is correct

Since, angles opposite to equal sides are equal

AB = AC

∠ABC = ∠ACB (i)

DB = DC

∠DBC = ∠DCB (ii)

Subtracting (ii) from (i),

∠ABC - ∠DBC = ∠ACB - ∠DCB

Hence, the assertion is true and also the reason gives the right explanation of the assertion.

∴ Option A is correct

In ΔBDC and ΔCEB,

∠DCB = ∠EBC (Given)

BC = CB (Common)

∠B = ∠C (AC = AB)

∠CEB = ∠BCE

∴ ΔBDC ≅ ΔCEB

BD = CE (By c.p.c.t.)

And, we know that the sum of two sides is always greater than the third side in any triangle.

But, (5 + 4) < 10

Hence, the reason is true, but the assertion is false.

∴ Option D is true

According to the question,

AB = AC

∠ACB = ∠ABC (i)

Now, ∠ACD >∠ACB =∠ABC (Side BC is produced to D)

And, In ∆ADC, side DC is produced to B

∠ACB>∠ADC (ii)

∠ABC>∠ADC

Now, using (i) and (ii),

AD >AB

Hence, the reason is wrong but the assertion is true.

∴ Option C is correct

The parts of the question are solved below:

a. Given: In △ABC, AB = AC and ∠A=50°

Thus, ∠B = ∠C

Now, ∠A + ∠B + ∠C = 180° (The angle sum property of triangle)

50 + 2∠B = 180°

2∠B = 130°

∠C = ∠B = 65°

b. As per the question,

Let the vertical angle be A and ∠ B = ∠ C

Now, ∠A + ∠B + ∠C = 180° (The angle sum property of triangle)

130 + 2∠B = 180°

2∠B = 50°

∠C = ∠B = 25°

c. We know that, the sum of three altitudes of a triangle ABC is less than its perimeter.

d. Here, ABCD is a square and EDC is a equilateral triangle.

∴ ED = CD = AB = BC = AD = EC

In ΔECB,

EC = BC

∠C = ∠B = x

∠ECD = 60° and ∠DCB = 90°

∠ECB = 60° + 90°

= 150°

Now, x + x + 150° = 180°

2x = 30°

x = 15°

∴∠EBC = 15°

∴ a = r, b = s, c = p, d = q

a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

a) Each angle of an equilateral triangle measures 60°

b) Medians of an equilateral triangle are equal

c) In a right triangle, the hypotenuse is the longest side

d) Drawing a △ABC with AB = 3cm, BC = 4cm and CA = 7cm is not possible.

We know that,

Each angle of an equilateral triangle is equal to 60^o also angles opposite to equal sides of a triangle are equal to each other

∴ Both assertion and reason are true and reason is the correct explanation of the assertion

Hence, option (A) is correct

We know that,

In any equilateral triangle all the angles are equal

Let the three angles of the triangle ∠A, ∠B and ∠C be x

∴ x + x + x = 180°

3x = 180°

x =

x = 60

Hence, ∠A = 60^o

It is given in the question that,

In triangle ABC, AB = AC

∠B = 65°

As ABC is an isosceles triangle

∴ ∠C = ∠B

∠C = 65°

Now, we now that sum of all angles of a triangle is 180^o

∴ ∠A + ∠B + ∠C = 180°

∠A + 65° + 65° = 180°

∠A + 130° = 180°

∠A = 180° – 130°

∠A = 50°

It is given in the question that,

In right triangle ABC,

∠B = 90^o

So, ∠A + ∠C = 90^o

∴ ∠A, ∠C < ∠B

Hence, the side opposite to ∠B is longest

Thus, AC is the longest side

It is given in the question that,

In triangle ABC, ∠B > ∠C

We know that, in a triangle side opposite to greater angle is longer

∴ AC is longer than AB

We know that,

The sum of two sides must be greater than the third side

In this case, we have

AB + BC = 5 + 4 = 9 cm

AC = 9 cm

∴ AC must be greater than the sum of AB and BC

Hence, the sum of two sides is not greater than the third side. So, cannot be constructed

From the figure, we have

∠AOD is the exterior angle

∴ ∠AOD + ∠AOB = 180^o

60^o + ∠AOB = 180^o

∠AOB = 180^o – 60^o

∠AOB = 120^o

Hence, the measure of each of the exterior angle of an equilateral triangle is 120^o

In a triangle let AC > AB

Then, along AC draw AD = AB and join BD

Proof: In Δ ABD,

∠ ABD = ∠ ADB (AB = AD) ….(i)

∠ ABD = ∠ 2 (angles opposite to equal sides) ….(ii)

Now, we know that the exterior angle of a triangle is greater than either of its opposite interior angles.

∴∠ 1 >∠ABD

∠1 > ∠2 ….(iii)

Now, from (ii)

∠2 > ∠3 ….(iv) (∠2 is an exterior angle)

Using (iii) and (iv),

∠1 > ∠3

BC > DC (side opposite to greater angle is longer)

BC > AC – AD

BC > AC – AB (since, AB = AD)

Hence, the difference of two sides is less than the third side of a triangle

It is given in the question that,

In right triangle ABC, ∠B = 90^o

Also D is the mid-point of AC

∴ AD = DC

∠ADB = ∠BDC (BD is the altitude)

BD = BD (Common)

So, by SAS congruence criterion

∴ ∆ADB ≅ ∆CDB

∠A = ∠C (CPCT)

As, ∠B = 90^o

So, by using angle sum property

∠A = ∠ABD = 45^o

Similarly, ∠BDC = 90^o (BD is the altitude)

∠C = 45^o

∠DBC = 45^o

∠ABD = 45^o

Now, by isosceles triangle property we have:

BD = CD and

BD = AD

AS, AD + DC = AC

BD + BD = AC

2BD = AC

BD =

Hence, proved

Let ABC be the triangle where D, E and F are the mid-points of BC, CA and AB respectively

As, we know that the sum of two sides of the triangle is greater than twice the median bisecting the third side

∴ AB + AC > 2AD

Similarly, BC + AC > 2CF

Also, BC + AB > 2BE

Now, by adding all these we get:

(AB + BC) + (BC + AC) + (BC + AB) > 2AD + 2CD + 2BE

2 (AB + BC + AC) > 2(AD + BE + CF)

∴ AB + BC + AC > AD + BE + CF

Hence, the perimeter of the triangle is greater than the sum of its medians

We know that,

A triangle can have two acute angles because the sum of two acute angles is always less than 180^o which satisfies the angle sum property of a triangle

Hence, option (A) is correct

It is given that,

BD is perpendicular to AC and CE is perpendicular to AB

Now, in ∆BDC and ∆CEB we have:

BE = CD (Given)

∠BEC = ∠CDB = 90^o

And, BC = BC (Common)

∴ By RHS congruence rule

∆BDC ≅ ∆CEB

BD = CE (By CPCT)

Hence, proved

It is given in the question that,

In ∆ABC,

AB = AC

We know that, angles opposite to equal sides are equal

∴ ∠ACB = ∠ABC

Now, in we have:

AC = AD

∠ADC = ∠ACD (The Angles opposite to equal sides are equal)

By using angle sum property in triangle BCD, we get:

∠ABC + ∠BCD + ∠ADC = 180^o

∠ACB + ∠ACB + ∠ACD + ∠ACD = 180^o

2 (∠ACB + ∠ACD) = 180^o

2 (∠BCD) = 180^o

∠BCD =

∠BCD = 90^o

Hence, proved

From the given figure,

In triangles DAC and CBD, we have:

AD = BC

AC = BD

DC = DC

So, by SSS congruence rule

∆ADC ≅ ∆BCD

∴ By Congruent parts of congruent triangles we have:

∠CAD = ∠CBD

∠ADC = ∠BCD

∠ACD = ∠BDC

Hence, proved

We have a triangle PQR where PS is the bisector of ∠ P

Now in ∆PQS and ∆PSR, we have:

PQ = PR (Given)

PS = PS (Common)

∠ QPS = ∠ PRS (As PS is the bisector of ∠ P)

∴ By SAS congruence rule

∆PQS ≅ ∆PSR

∠ Q = ∠ R (By Congruent parts of congruent triangles)

Hence, it is proved that the angles opposite to equal sides of a triangle are equal

From the given figure, we have:

(i) In ∆ABO and ∆ACO

AB = AC (Given)

AO = AO (Common)

∠ ABO = ∠ ACO

∴ By SAS congruence rule

OB = OB (By CPCT)

(ii) As, By SAS congruence rule

∆ABO ≅ ∆ACO

∴ ∠ OAB = ∠ OAC (By Congruent parts of congruent triangles)

Hence, proved

It is given in the question that,

l is the straight line and A is a point that is not lying on l

AB is perpendicular to line l and C is the point on l

As, ∠ B = 90^o

So in ∆ABC, we have:

∠ A + ∠ B + ∠ C = 180^o

∠ A + ∠ B = 90^o

∴ ∠ C < 90^o

∠ C < ∠ B

AB < AC

As C is that point which can lie anywhere on l

∴ AB is the shortest line segment drawn from A to l

Hence, proved

From the given figure in the question, we have

In ∆ABD, we have:

AB + BD > AD

Similarly, in

AC + CD > AD

Adding both expressions, we get:

AB + AC + BD + CD > AD + AD

AB + AC + BD + DC > 2AD

AB + AC + BC > 2AD

∴ Assertion and reason both are true and reason is the correct explanation of the assertion

Hence, option (A) is correct

a) In ∆ABC, ∠ A=70°

As AB = AC and we know that angles opposite to equal sides are equal

∴ In triangle ABC,

∠ A + ∠ B + ∠ C = 180^o

70^o + 2∠ C = 180^o

2∠ C = 180^o – 70^o

∠ C =

∴ ∠ C = 55^o

(b) We know that,

Angles opposite to equal sides are equal

It is given that, vertical angle of the isosceles triangle = 120^o

Let the base angle be x

∴ 120° + x + x = 180°

120° + 2x = 180°

2x = 180° – 120°

2x = 60°

x =

x = 30°

Hence, each base angle of the isosceles triangle is equal to 30^o

(c) We know that,

The sum of the three medians of the triangle is always less than the perimeter

(d) We know that,

In a triangle the sum of any two sides is always greater than the third side

Hence, the correct match is as follows:

(a) – (s)

(b) – (r)

(c) – (p)

(d) – (q)

It is given in the question that,

PQ > PR

And, QS and RS are the bisectors of ∠ Q and ∠ R

We have, angle opposite to the longer side is greater

∴ PQ > PR

∠ R > ∠ Q

∠ SRQ > ∠ RQS

SQ > SR

Hence, proved

We will have to make the following construction in the given figure:

Produce CB to D in such a way that BD=BC and join AD.

Now, in ∆ABC and ∆ABD,

BC=BD (constructed)

AB=AB (common)

∠ABC=∠ABD (each 90°)

∴ by S.A.S.

∆ABC ≅ ∆ABD

∠CAB=∠DAB and AC=AD (by c.p.c.t.)

∴∠CAD=∠CAB+∠BAD

=x°+x°

=2x°

But, AC=AD

∠ACD=∠ADB=2x°

∴ ∆ACD is equilateral triangle.

AC=CD

AC=2BC

Hence, proved

Following construction is to be made in the given figure.

Extend QS to meet PR at T.

Now, in ∆ PQT,

PQ+PT>QT (sum of two sides is greater than the third side in a triangle)

PQ+PT>SQ+ST (i)

Now, In ∆ STR,

ST+TR>SR (ii)(sum of two sides is greater than the third side in a triangle)

Now, adding (i) and (ii),

PQ+PT+ST+TR>SQ+ST+SR

PQ+PT+TR>SQ+SR

PQ+PR>SQ+SR

SQ+SR<PQ+PR

Hence, proved

Here, ABCD is a quadrilateral and AC and BD are its diagonals.

Now, As we that, sum of two sides of a triangle is greater than the third side.

∴ In Δ ACB,

AB + BC > AC (i)

In Δ BDC,

CD + BC > BD (ii)

In Δ BAD,

AB + AD>BD (iii)

In Δ ACD,

AD + DC > AC (iv)

Now, adding (i), (ii), (iii) and (iv):

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

2AB + 2BC + 2CD + 2AD > 2AC + 2BD

Thus, AB + BC + CD + AD > AC + BD

Hence, proved