# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 9 - Congruence of Triangles and Inequalities in a Triangle

## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise MCQ

Which of the following is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSS

Correct option: (a)

SSA is not a criterion for congruence of triangles.

If AB = QR, BC = RP and CA = PQ, then which of the following holds?

(a) ∆ABC ≅ ∆PQR

(b) ∆CBA ≅ ∆PQR

(c) ∆CAB ≅ ∆PQR

(d) ∆BCA ≅ ∆PQR

Correct option: (c)

If ∆ABC ≅ ∆PQR then which of the following is not true?

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQ

In Δ ABC, AB = AC and ∠B = 50°. Then, ∠A = ?

(a) 40°

(b) 50°

(c) 80°

(d) 130°

Correct option: (c)

In ΔABC,

AB = AC

⇒ ∠C = ∠B (angles opposite to equal sides are equal)

⇒ ∠C = 50°

Now, ∠A + ∠B + ∠C = 180°

⇒ ∠A + 50° + 50° = 180°

⇒ ∠A = 80°

In Δ ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

(a) 50°

(b) 40°

(c) 100°

(d) 80°

Correct option: (a)

In ΔABC,

BC = AB

⇒ ∠A = ∠C (angles opposite to equal sides are equal)

Now, ∠A + ∠B + ∠C = 180°

⇒ ∠A + 80° + ∠A = 180°

⇒ 2∠A = 100°

⇒ ∠A = 50°

In ΔABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 2.5 cm

Correct option: (a)

In ΔABC,

∠C = ∠A

⇒ AB = BC (sides opposite to equal angles are equal)

⇒ AB = 4 cm

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be

(a) 6 cm

(b) 6.5 cm

(c) 5.5 cm

(d) 6.3 cm

Correct option: (b)

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

Since, 4 cm + 2.5 cm = 6.5 cm

The length of third side of a triangle cannot be 6.5 cm.

In ΔABC, if ∠C > ∠B, then

(a) BC > AC

(b) AB > AC

(c) AB < AC

(d) BC < AC

Correct option: (b)

We know that in a triangle, the greater angle has the longer side opposite to it.

In ΔABC,

∠C > ∠B

⇒ AB >AC

It is given that ∆ABC ≅ ∆FDE in which AB
= 5 cm, ∠B = 40^{o}, ∠A = 80^{o}
and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60^{o}

(b) ∠E = 60^{o}

(c) ∠F = 60^{o}

(d) ∠D = 80^{o}

In ∆ABC, ∠A = 40^{o}
and ∠B = 60^{o}.
Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be determined

In the given figure AB > AC. Then, which of the following is true?

(a) AB < AD

(b) AB = AD

(c) AB > AD

(d) Cannot be determined

Correct option: (c)

In the given figure AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then

(a) OB = OC

(b) OB > OC

(c) OB < OC

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?

(a) 1 :1

(b) 2 : 1

(c) 1 :2

(d) None of these

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

(a) Equilateral

(b) Isosceles

(c) Scalene

(d) Right-angled

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have

(a) ∠A = ∠D

(b) ∠B = ∠E

(c) ∠C = ∠F

(d) None of these

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have

(a) AB = DF

(b) AC = DE

(c) BC = EF

(d) ∠A = ∠D

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are

(a) Isosceles but not congruent

(b) Isosceles but congruent

(c) Congruent but not isosceles

(d) Neither congruent nor isosceles

Which is true ?

(a) A triangle can have two right angles.

(b) A triangle can have two obtuse angles.

(c) A triangle can have two acute angles.

(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Fill in the blanks with

(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 sides

(d) (Sum of any two sides of a triangle)….. (twice the median to the 3^{rd} side)

(e) (Perimeter of a triangle)……(sum of its medians)

Fill in the blanks

(a) Each angle of an equilateral triangles measures …….

(b) Medians of an equilateral triangle are ……….

(c) In a right triangle the hypotenuse is the ….. side

(d) Drawing a ∆ABC with AB = 3cm, BC= 4 cm and CA = 7 cm is ……..

## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise Ex. 9B

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

5 cm, 4 cm, 9 cm

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 5 cm and 4 cm, is not greater than the third side, 9 cm.

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

8 cm, 7 cm, 4 cm

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

10 cm, 5 cm, 6 cm

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

2.5 cm, 5 cm, 7 cm

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side.

3 cm, 4 cm, 8 cm

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

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ 50° + 60° + ∠C = 180°

⇒ ∠C = 70°

Thus, we have

∠A < ∠B < ∠C

⇒ BC < AC < AB

Hence, the longest side is AB and the shortest side is BC.

In ΔABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ 100° + ∠B + 50° = 180°

⇒ ∠B = 30°

Thus, we have

∠B < ∠C < ∠A

⇒ AC < AB < BC

Hence, the shortest side is AC.

In _{}ABC, if _{}A = 90^{o}, which is the longest side?

_{}

In _{}ABC, if _{}A = _{}B = 45^{o}, name the longest side.

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In _{}ABC, side AB is produced to D such that BD = BC. If _{}B = 60^{o} and _{}A = 70^{o}, prove that (i) AD > CD and (ii) AD > AC.

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In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

In ΔAOB,

∠B < ∠A

⇒ AO < BO ….(i)

In ΔCOD,

∠C < ∠D

⇒ DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒ AD < BC

AB and CD are respectively the smallest and largest sides of quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Construction: Join AC and BD.

In ΔABC,

BC > AB

⇒ ∠BAC > ∠ACB ….(i)

In ΔACD,

CD > AD

⇒ ∠CAD > ∠ACD ….(ii)

Adding (i) and (ii), we get

∠BAC + ∠CAD > ∠ACB + ∠ACD

⇒ ∠A > ∠C

In ΔADB,

AD > AB

⇒ ∠ABD > ∠ADB ….(iii)

In ΔBDC,

CD > BC

⇒ ∠CBD > ∠BDC ….(iv)

Adding (iii) and (iv), we get

∠ABD + ∠CBD > ∠ADB + ∠BDC

⇒ ∠B > ∠D

In a quadrilateral ABCD, show that

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

In ΔABC,

AB + BC > AC ….(i)

In ΔACD,

DA + CD > AC ….(ii)

In ΔADB,

DA + AB > BD ….(iii)

In ΔBDC,

BC + CD > BD ….(iv)

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

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) < 2(BD + AC).

In ΔAOB,

AO + BO > AB ….(i)

In ΔBOC,

BO + CO > BC ….(ii)

In ΔCOD,

CO + DO > CD ….(iii)

In ΔAOD,

DO + AO > DA ….(iv)

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

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

⇒ 2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

⇒ 2AC + 2BD > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2(AC + BD)

In _{}ABC, _{}B = 35^{o}, _{}C = 65^{o} and the bisector of _{}BAC meets BC in X. Arrange AX, BX and CX in descending order.

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In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

In ΔPQR,

PQ > PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SR

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.

In ΔABC,

AB = AC

⇒ ∠ABC = ∠ACB ….(i)

Now, ∠ABC = ∠ABD + ∠DBC

⇒ ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC [From (i)]

⇒ ∠DCB > ∠DBC

⇒ BD > CD

i.e. CD < BD

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than of a right angle.

Let PQR be the required triangle.

Let PR be the longest side.

Then, PR > PQ

⇒ ∠Q > ∠R ….(i)

Also, PR > QR

⇒ ∠Q > ∠P ….(ii)

Adding (i) and (ii), we get

2∠Q > ∠R + ∠P

⇒ 2∠Q + ∠Q > ∠P + ∠Q + ∠R (adding ∠Q to both sides)

⇒ 3∠Q > 180°

⇒ ∠Q > 60°

In the given figure, prove that CD + DA + AB > BC

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AC + AB > BC ….(ii)

Adding (i) and (ii), we get

CD + DA + AC + AB > AC + BC

Subtracting AC from both sides, we get

CD + DA + AB > BC

In the given figure, prove that

CD + DA + AB + BC > 2AC.

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AB + BC > AC ….(ii)

Adding (i) and (ii), we get

CD + DA + AB + BC > AC + AC

⇒ CD + DA + AB + BC > 2AC

If O is a point within _{}ABC, show that:

AB + AC > OB + OC

Given : ABC is a triangle and O is appoint insideit.

To Prove : (i) AB+AC > OB +OC

If O is a point within _{}ABC, show that:

AB + BC + CA > OA + OB + OC

AB+BC+CA > OA+OB+OC

If O is a point within _{}ABC, show that:

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

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

Proof:

(i)In_{}ABC,

AB+AC>BC.(i)

And in , _{}OBC,

OB+OC>BC.(ii)

Subtracting (i) from (i) we get

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

i.e.AB+AC>OB+OC

(ii)AB+AC> OB+OC[proved in (i)]

Similarly,AB+BC > OA+OC

AndAC+BC> OA +OB

Addingboth sides of these three inequalities, we get

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

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

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii)In_{}OAB

OA+OB > AB(i)

In_{}OBC,

OB+OC > BC(ii)

And, in _{}OCA,

OC+OA>CA

Adding (i), (ii) and (iii)we get

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

i.e2(OA+OB+OC) > AB+BC+CA

_{}OA+OB+OC> (AB+BC+CA)

In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Construction: Mark a point S on BC such that BD = SD. Join AS.

In ΔADB and ΔADS,

BD = SD (by construction)

∠ADB = ∠ADS (Each equal to 90°)

AD = AD (common)

∴ ΔADB ≅ ΔADS (by SAS congruence criterion)

⇒ AB = AS (c.p.c.t.)

Now, in ΔABS,

AB = AS

⇒ ∠ASB = ∠ABS ….(i)(angles opposite to equal sides are equal)

In ΔACS,

∠ASB > ∠ACS ….(ii)

From (i) and (ii), we have

∠ABS > ∠ACS

⇒ ∠ABC > ∠ACB

⇒ AC > AB

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

In ΔABC,

AB + AC > BC

⇒ AB + AC >BD + DC

⇒ AB + AC >BD + DE ….(i) [since CD = DE]

In ΔBED,

BD + DE > BE ….(ii)

From (i) and (ii), we have

AB + AC > BE

## Chapter 9 - Congruence of Triangles and Inequalities in a Triangle Exercise Ex. 9A

In the given figure, AB ∥ CD and O is the midpoint of AD.

Show that (i) Δ AOB ≅ Δ DOC (ii) O is the midpoint of BC.

(i) In ΔAOB and ΔDOC,

∠BAO = ∠CDO (AB ∥ CD, alternate angles)

AO = DO (O is the mid-point of AD)

∠AOB = ∠DOC (vertically opposite angles)

∴ ΔAOB ≅ ΔDOC (by ASA congruence criterion)

(ii) Since ΔAOB ≅ ΔDOC,

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

⇒ O is the mid-point of BC.

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisect AB.

In ΔAOD and ΔBOC,

∠AOD = ∠BOC (vertically opposite angles)

∠DAO = ∠CBO (Each 90°)

AD = BC (given)

∴ ΔAOD ≅ BOC (by AAS congruence criterion)

⇒ AO = BO (c.p.c.t.)

⇒ CD bisects AB.

In the given figure, two parallels lines l and m are intersected by two parallels lines p and q. Show that Δ ABC ≅ Δ CDA.

In ΔABC and ΔCDA

∠BAC = ∠DCA (alternate interior angles for p ∥ q)

AC = CA (common)

∠BCA = ∠DAC (alternate interior angles for l ∥ m)

∴ ΔABC ≅ ΔCDA (by ASA congruence rule)

AD is an altitude of an isosceles ΔABC in which AB = AC.

Show that (i) AD bisects BC, (ii) AD bisects ∠A.

(i) In ΔBAD and ΔCAD

∠ADB = ∠ADC (Each 90° as AD is an altitude)

AB = AC (given)

AD = AD (common)

∴ ΔBAD ≅ ΔCAD (by RHS Congruence criterion)

⇒ BD = CD (c.p.c.t.)

Hence AD bisects BC.

(ii) Also, ∠BAD = ∠CAD (c.p.c.t.)

Hence, AD bisects ∠A.

In the given figure, BE and CF are two equal altitudes of ΔABC.

Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90°)

BE = CF (given)

∠BAE = ∠CAF (common ∠A)

∴ ΔABE ≅ ACF (by ASA congruence criterion)

(ii) Since ΔABE ≅ ΔACF,

AB = AC (c.p.c.t.)

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABE ≅ ΔACE

(iii) AE bisects ∠A as well as ∠D

(iv) AE is the perpendicular bisector of BC.

(i) In ΔABD and ΔACD,

AB = AC (equal sides of isosceles ΔABC)

DB = DC (equal sides of isosceles ΔDBC)

AD = AD (common)

∴ ΔABD ≅ ACD (by SSS congruence criterion)

(ii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE ….(1)

Now, in ΔABE and ΔACE

AB = AC (equal sides of isosceles ΔABC)

∠BAE = ∠CAE [From (1)]

AE = AE (common)

∴ ΔABE ≅ ACE (by SAS congruence criterion)

(iii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE

Thus, AE bisects ∠A.

In ΔBDE and ΔCDE,

BD = CD (equal sides of isosceles ΔABC)

BE = CE (c.p.c.t. since ΔABE ≅ ACE)

DE = DE (common)

∴ ΔBDE ≅ CDE (by SSS congruence criterion)

⇒ ∠BDE = ∠CDE (c.p.c.t.)

Thus, DE bisects ∠D, i.e., AE bisects ∠D.

Hence, AE bisects ∠A as well as ∠D.

(iv) Since ΔBDE ≅ ΔCDE,

BE = CE and ∠BED = ∠CED (c.p.c.t.)

⇒ BE = CE and ∠BED = ∠CED = 90° (since ∠BED and ∠CED form a linear pair)

⇒ DE is the perpendicular bisector of BC.

⇒ AE is the perpendicular bisector of BC.

In the given figure, if x = y and AB = CB, then prove that AE = CD.

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

(i) In ΔAPB and ΔAQB,

∠APB = ∠AQC (Each 90°)

∠BAP = ∠BAQ (line l is the bisector of ∠A)

AB = AB (common)

∴ ΔAPB ≅ AQB (by AAS congruence criterion)

(ii) Since ΔAPB ≅ ΔAQB,

BP = BQ (c.p.c.t.)

ABCD is a quadrilateral such that diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.

In ΔABC and ΔADC,

∠BAC = ∠DAC (AC bisects ∠A)

AC = AC (common)

∠BCA = ∠DCA (AC bisects ∠C)

∴ ΔABC ≅ ADC (by ASA congruence criterion)

⇒ AB = AD and CB = CD (c.p.c.t.)

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersect the side AB at D. Prove that AC + AD = BC.

Construction: Draw DE ⊥ BC.

In ΔDAC and ΔDEC,

∠DAC = ∠DEC (Each 90°)

∠DCA = ∠DCE (CD bisects ∠C)

CD = CD (common)

∴ ΔDAC ≅ ΔDEC (by AAS congruence criterion)

⇒ DA = DE (c.p.c.t.) ….(i)

and AC = EC (c.p.c.t.) ….(ii)

Given, AB = AC

⇒ ∠B = ∠C (angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180°

⇒ 90° + ∠B + ∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

In ΔBED,

∠BDE + ∠B = 90° (since ∠BED = 90°)

⇒ ∠BDE + 45° = 90°

⇒ ∠BDE = 45°

⇒ ∠BDE = ∠DBE = 45°

⇒ DE = BE ….(iii)

From (i) and (iii),

DA = DE = BE ….(iv)

Now, BC = BE + EC

⇒ BC = DA + AC [From (ii) and (iv)

⇒ AC + AD = BC

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX (ii) AX = BX.

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In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

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In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

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In the given figure, ABCD is square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

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In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

ΔOAB is an equilateral triangle.

⇒ ∠OAB = ∠OBA = AOB = 60°

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90°

Now, ∠A = ∠DAO + ∠OAB

⇒ 90° = ∠DAO + 60°

⇒ ∠DAO = 90° - 60° = 30°

Similarly, ∠CBO = 30°

In ΔOAD and ΔOBC,

AD = BC (sides of a square ABCD)

∠DAO = ∠CBO = 30°

OA = OB (sides of an equilateral ΔOAB)

∴ ΔOAD ≅ ΔOBC (by SAS congruence criterion)

⇒ OD = OC (c.p.c.t.)

Hence, ΔOCD is an isosceles triangle.

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of _{}ABC such that AX = AY. Prove that CX = BY.

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In _{}ABC, D is the midpoint of BC. If DL _{}AB and DM _{}AC such that DL = DM, prove that AB = AC.

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In _{}ABC, AB = AC and the bisectors of _{B} and _{}C meet at a point O. Prove that BO = CO and the ray AO is the bisector _{}A.

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Construction: Join AN and BN.

In ΔANM and ΔBNM

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (Each 90°)

MN = MN (common)

∴ ΔANM ≅ ΔBNM (by SAS congruence criterion)

⇒ AN = BN (c.p.c.t.) ….(i)

And, ∠ANM = ∠BNM (c.p.c.t.)

⇒ 90° - ∠ANM = 90° - ∠BNM

⇒ ∠AND = ∠BNC ….(ii)

In ΔAND and DBNC,

AN = BN [From (i)]

∠AND = ∠BNC [From (ii)]

DN = CN (N is the mid-point of DC)

∴ ΔAND ≅ ΔBNC (by SAS congruence criterion)

⇒ AD = BC (c.p.c.t.)

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

Now, by exterior angle property,

∠MOC = ∠OBC + ∠OCB

⇒ ∠MOC = 2∠OBC [From (i)]

⇒ ∠MOC = ∠ABC (OB is the bisector of ∠ABC)

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

In ΔBOC, by angle sum property,

∠BOC + ∠OBC + ∠OCB = 180°

⇒ ∠BOC + 2∠OBC = 180° [From (i)]

⇒ ∠BOC + ∠ABC = 180°

⇒ ∠BOC + (180° - ∠ABP) = 180° (∠ABC and ∠ABP form a linear pair)

⇒ ∠BOC + 180° - ∠ABP = 180°

⇒ ∠BOC - ∠ABP = 0

⇒ ∠BOC = ∠ABP

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.

AB ∥ PQ and BP is a transversal.

⇒ ∠ABP = ∠BPQ (alternate angles) ….(i)

BP is the bisector of ∠ABC.

⇒ ∠ABP = ∠PBC

⇒ ∠ABP = ∠PBQ ….(ii)

From (i) and (ii), we have

∠BPQ = ∠PBQ

⇒ PQ = BQ (sides opposite to equal angles are equal)

⇒ ΔBPQ is an isosceles triangle.

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

To prove that the image is as far behind the mirror as the object is in front of the mirror, we need to prove that AT = BT.

We know that angle of incidence = angle of reflection.

⇒ ∠ACN = ∠DCN ….(i)

AB ∥ CN and AC is the transversal.

⇒ ∠TAC = ∠ACN (alternate angles) ….(ii)

Also, AB ∥ CN and BD is the transversal.

⇒ ∠TBC = ∠DCN (corresponding angles) ….(iii)

From (i), (ii) and (iii),

∠TAC = ∠TBC ….(iv)

In ΔACT and ΔBCT,

∠TAC = ∠TBC [From (iv)]

∠ATC = ∠BTC (Each 90°)

CT = CT (common)

∴ ΔACT ≅ ΔBCT (by AAS congruence criterion)

⇒ AT = BT (c.p.c.t.)

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

_{Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM. }

_{}

_{}

In a ΔABC, D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.

D is the mid-point of AC.

⇒ AD = CD =

Given, BD =

⇒ AD = CD = BD

Consider AD = BD

⇒ ∠BAD = ∠ABD (i)(angles opposite to equal sides are equal)

Consider CD = BD

⇒ ∠BCD = ∠CBD (ii)(angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠ABC + ∠BAC + ∠BCA = 180°

⇒ ∠ABC + ∠BAD + ∠BCD = 180°

⇒ ∠ABC + ∠ABD + ∠CBD = 180° [From (i) and (ii)]

⇒ ∠ABC + ∠ABC = 180°

⇒ 2∠ABC = 180°

⇒ ∠ABC = 90°

Hence, ∠ABC is a right angle.

"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent." Is the statement true? Why?

The given statement is not true.

Two triangles are congruent if two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle.

"If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent." Is the statement true? Why?

The given statement is not true.

Two triangles are congruent if two angles and the included side of one triangle are equal to corresponding two angles and the included angle of another triangle.

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