RS Agarwal Solution | Class 9th | Chapter-9 | Congruence of Triangles and Inequalities in a Triangle | Edugrown

Exercise MCQ

Question 1

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

(a) SSA

(b) SAS

(c) ASA

(d)SSSSolution 1

Correct option: (a)

SSA is not a criterion for congruence of triangles.Question 2

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 ≅ ∆PQRSolution 2

Correct option: (c)

Question 3

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

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQSolution 3

Question 4

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

(a) 40° 

(b) 50° 

(c) 80° 

(d) 130° Solution 4

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° Question 5

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

(a) 50° 

(b) 40° 

(c) 100° 

(d) 80° Solution 5

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°  Question 6

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 cmSolution 6

Correct option: (a)

In ΔABC,

∠C = ∠A

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

⇒ AB = 4 cm Question 7

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 cmSolution 7

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. Question 8

In ΔABC, if ∠C > ∠B, then

(a) BC > AC

(b) AB > AC

(c) AB < AC

(d) BC < ACSolution 8

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  Question 9

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^oSolution 9

Question 10

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 determinedSolution 10

Question 11

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

Solution 11

Correct option: (c)

Question 12

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

Solution 12

Question 13

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 

Solution 13

Question 14

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-angledSolution 14

Question 15

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

Solution 15

Question 16

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

Solution 16

Question 17

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

Solution 17

Question 18

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.

Solution 18

Question 19

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)Solution 19

Question 20

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 ……..Solution 20

Exercise Ex. 9B

Question 1(i)

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 cmSolution 1(i)

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. Question 1(ii)

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 cmSolution 1(ii)

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. Question 1(iii)

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 cmSolution 1(iii)

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. Question 1(iv)

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

2.5 cm, 5 cm, 7 cmSolution 1(iv)

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. Question 1(v)

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

3 cm, 4 cm, 8 cmSolution 1(v)

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. Question 2

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.Solution 2

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. Question 3(iii)

In ΔABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?Solution 3(iii)

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. Question 3(i)

In ABC, if A = 90^o, which is the longest side?Solution 3(i)

Question 3(ii)

In ABC, if A = B = 45^o, name the longest side.Solution 3(ii)

Question 4

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.

Solution 4

Question 5

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

Solution 5

In ΔAOB,

∠B < ∠A

⇒ AO < BO ….(i)

In ΔCOD,

∠C < ∠D

⇒ DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒ AD < BC Question 6

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

Solution 6

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 Question 7

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) > (AC + BD).Solution 7

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 Question 8

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) < 2(BD + AC).Solution 8

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) Question 9

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.

Solution 9

Question 10

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

Solution 10

In ΔPQR,

PQ > PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SRQuestion 11

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.Solution 11

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 Question 12

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than  of a right angle.Solution 12

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° 

Question 13(i)

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

Solution 13(i)

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 Question 13(ii)

In the given figure, prove that

CD + DA + AB + BC > 2AC.

Solution 13(ii)

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 Question 14(i)

If O is a point within ABC, show that:

AB + AC > OB + OCSolution 14(i)

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

To Prove : (i) AB+AC > OB +OCQuestion 14(ii)

If O is a point within ABC, show that:

AB + BC + CA > OA + OB + OCSolution 14(ii)

AB+BC+CA > OA+OB+OCQuestion 14(iii)

If O is a point within ABC, show that:

OA + OB + OC > (AB + BC + CA)Solution 14(iii)

OA+OB+OC> (AB+BC+CA)

Proof:

(i)InABC,

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)InOAB

OA+OB > AB(i)

InOBC,

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)Question 15

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

Solution 15

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 > ABQuestion 16

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.

Solution 16

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

Exercise Ex. 9A

Question 1

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.

Solution 1

(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.Question 2

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

Solution 2

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. Question 3

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.

Solution 3

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)Question 4

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

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

Solution 4

(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.Question 5

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

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

Solution 5

(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.)Question 6

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.

Solution 6

(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.Question 7

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

Solution 7

Question 8

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.

Solution 8

(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.) Question 9

ABCD is a quadrilateral such that diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.Solution 9

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.) Question 10

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.Solution 10

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 = BCQuestion 11

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

Solution 11

Question 12

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.

Solution 12

Question 13

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.

Solution 13

Question 14

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.

Solution 14

Question 15

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.

Solution 15

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.Question 16

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.

Solution 16

Question 17

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

Solution 17

Question 18

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.

Solution 18

Question 19

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.Solution 19

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.)Question 20

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.Solution 20

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)  Question 21

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.Solution 21

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 Question 22

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.Solution 22

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.Question 23

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.

Solution 23

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.)Question 24

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

Solution 24

_{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.}

Question 25

In a ΔABC, D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.Solution 25

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.Question 26

“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?Solution 26

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. Question 27

“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?Solution 27

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.