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Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

Isosceles Triangle Exercise Ex. 10(A)

Solution 1

In

BAC + ACB + ABC = 1800

480 + ACB + ABC = 1800

But ACB = ABC [AB = AC]

2ABC = 1800 - 480

2ABC = 1320

ABC = 660 = ACB ……(i)

ACB = 660

ACD + DCB = 660

180 + DCB = 660

DCB = 480 ………(ii)

Now, In

DBC = 660 [From (i), Since ABC = DBC]

DCB = 480 [From (ii)]

BDC = 1800 - 480 - 660

BDC = 660

Since BDC = DBC

Therefore, BC = CD

Equal angles have equal sides opposite to them.

Solution 2

Given: ACE = 1300; AD = BD = CD

Proof:

(i)

(ii)

begin mathsize 11px style angle ADC space equals space angle ABD space plus space angle DAB space open square brackets Exterior space angle space is space equal space to space sum space of space opp. space interior space angles close square brackets
But space AD space equals space BD
therefore space angle DAB space equals space angle ABD
rightwards double arrow 80 to the power of ring operator space equals space angle ABD space plus space angle ABD
rightwards double arrow 2 angle ABD space equals space 80 to the power of ring operator
rightwards double arrow angle ABD space equals space 40 to the power of ring operator minus angle DAB space........ left parenthesis ii right parenthesis end style

(iii)

Solution 3

(i)

(ii)

Given space DE vertical line vertical line BC space and space DC space is space the space transversal.
rightwards double arrow angle CDE equals angle DCB equals 52 degree....... left parenthesis ii right parenthesis
Also comma space angle ECB equals 64 degree......... left square bracket From space left parenthesis straight i right parenthesis right square bracket
But comma
angle ECB equals angle DCE plus angle DCB
rightwards double arrow 64 degree equals angle DCE plus 52 degree
rightwards double arrow angle DCE equals 64 degree minus 52 degree
rightwards double arrow angle DCE equals 12 degree

Solution 4

(i) Let the triangle be ABC and the altitude be AD.

 

(ii) Let triangle be ABC and altitude be AD.

Solution 5

Let ABO =OBC = x and ACO = OCB = y

Now,

Solution 6

Given:

(i) We know that the sum of the measure of all the angles of a quadrilateral is 360o.

In quad. PQNL,

(ii)

Solution 7

Now,

Solution 8

Let us name the figure as following:

For x:

Solution 9

Therefore,

AD=DC

and AB = BC

Substituting the value of x from (i)

Putting y = 3 in (i)

x = 3 + 1

x = 4

Solution 10

Let P and Q be the points as shown below:

Given:

Solution 11

Now,

Solution 12

Solution 13

Let

Given: AB = AC

[Angles opp. to equal sides are equal]

Solution 14

Now,

BP is the bisector of

Solution 15

Let PBC = PCB = x

In the right angled triangle ABC,

and

Therefore in the triangle ABP;

Hence,

PA = PB [sides opp. to equal angles are equal]

Solution 16

  

 

Solution 17

 

  

Solution 18

Solution 19

(i)  In ΔABC, let the altitude AD bisects BAC.

Then we have to prove that the ΔABC is isosceles.

  

In triangles ADB and ADC,

BAD = CAD (AD is bisector of BAC)

AD = AD (common)

ADB = ADC (Each equal to 90°)

ΔADB ΔADC (by ASA congruence criterion)

AB = AC (cpct)

Hence, ΔABC is an isosceles.

(ii)  In Δ ABC, the bisector of BAC is perpendicular to the base BC. We have to prove that the ΔABC is isosceles.

  

 In triangles ADB and ADC,

BAD = CAD (AD is bisector of BAC)

AD = AD (common)

ADB = ADC (Each equal to 90°)

ΔADB ΔADC (by ASA congruence criterion)

AB = AC (cpct)

Hence, ΔABC is an isosceles.

Solution 20

  

In ΔABC,

AB = BC (given)

BCA = BAC (Angles opposite to equal sides are equal)

BCD = BAE ….(i)

Given, AD = EC

AD + DE = EC + DE (Adding DE on both sides)

AE = CD ….(ii)

Now, in triangles ABE and CBD,

AB = BC (given)

BAE = BCD [From (i)]

AE = CD [From (ii)]

ΔABE ΔCBD

BE = BD (cpct)

Isosceles Triangle Exercise Ex. 10(B)

Solution 1

Const: AB is produced to D and AC is produced to E so that exterior angles and is formed.

Since angle B and angle C are acute they cannot be right angles or obtuse angles.

Now,

Therefore, exterior angles formed are obtuse and equal.

Solution 2

Const: Join AD.

(i)

(ii) We have already proved that

Therefore,BP = CQ[cpct]

Now,

AB = AC[Given]

AB - BP = AC - CQ

AP = AQ

(iii)

Hence, AD bisects angle A.

Solution 3

(i)

In space increment AEB space and space increment AFC comma
angle straight A equals angle straight A space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Common right square bracket
angle AEB equals angle AFC space equals 90 degree space space left square bracket Given : space BE perpendicular AC right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Given : CF perpendicular AB right square bracket
AB equals AC space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Given right square bracket
rightwards double arrow increment AEB approximately equal to increment AFC space space space space space space space space space space left square bracket AAS right square bracket
therefore BE equals CF space space space space space space space space space space space space space space space space space space space space space space space space left square bracket cpct right square bracket

(ii)Since

angle ABE equals angle AFC space
therefore AF equals AE space space space space space space space space space space space space space space space space space space space space left square bracket congruent space angles space of space congruent space triangles right square bracket

Solution 4

Const: Join CD.

Adding (i) and (ii)

Solution 5(i)

  

Solution 5(ii)

Given:

AD is the angle bisector of ∠BAC, hence ∠BAD = ∠DAC

Also, AD bisects BC, hence BD = DC.

To prove: ∆ABC is isosceles, i.e. AB = AC

Proof:

In ∆ABC

BD = DC … (given)

Now by angle bisector theorem

Solution 6

Solution 7

DBC = ECB = 90o[Given]

DBC = ECB …….(ii)

Subtracting (i) from (ii)

Solution 8

DA is produced to meet BC in L.

Subtracting (i) from (ii)

From (iii), (iv) and (v)

From (vi) and (vii)

Now,

Solution 9

In ABC, we have AB = AC

B = C [angles opposite to equal sides are equal]

Now,

In ABO and ACO,

AB = AC [Given]

OBC = OCB [From (i)]

OB = OC [From (ii)]

Therefore, AO bisects BAC.

Solution 10

Solution 11

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

Solution 12

Since AE || BC and DAB is the transversal

Since AE || BC and AC is the transversal

angle CAE equals angle ACB equals angle straight C space space space space space space space space space space space space space space space left square bracket Alternate space Angles right square bracket

But AE bisects

AB = AC[Sides opposite to equal angles are equal]

Solution 13

AB = BC = CA…….(i) [Given]

AP = BQ = CR…….(ii) [Given]

Subtracting (ii) from (i)

AB - AP = BC - BQ = CA - CR

BP = CQ = AR …………(iii)

……..(iv) [angles opp. to equal sides are equal]

From (v) and (vi)

PQ = QR = PR

Therefore, PQR is an equilateral triangle.

Solution 14

In ABE and ACF,

A = A[Common]

AEB = AFC = 900[Given: BE AC; CF AB]

BE = CF[Given]

Therefore, ABC is an isosceles triangle.

Solution 15

AL is bisector of angle A. Let D is any point on AL. From D, a straight line DE is drawn parallel to AC.

DE || AC [Given]

ADE = DAC….(i) [Alternate angles]

DAC = DAE…….(ii) [AL is bisector of A]

From (i) and (ii)

ADE = DAE

AE = ED [Sides opposite to equal angles are equal]

Therefore, AED is an isosceles triangle.

Solution 16

(i)

In ABC,

AB = AC

 

AP = AQ …….(i)[ Since P and Q are mid - points]

In BCA,

PR = [PR is line joining the mid - points of AB and BC]

PR = AQ……..(ii)

In CAB,

QR = [QR is line joining the mid - points of AC and BC]

QR = AP……(iii)

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

PR = QR

 

(ii)

AB = AC

B = C

Also,

In BPC and CQB,

BP = CQ

B = C

BC = BC

Therefore, ΔBPCapproximately equal toΔCQB   [SAS]

BP = CP

Solution 17

(i) In ACB,

AC = AC[Given]

ABC = ACB …….(i)[angles opposite to equal sides are equal]

ACD + ACB = 1800 …….(ii)[DCB is a straight line]

ABC + CBE = 1800 ……..(iii)[ABE is a straight line]

Equating (ii) and (iii)

ACD + ACB = ABC + CBE

ACD + ACB = ACB + CBE[From (i)]

ACD = CBE

(ii)

Solution 18

AB is produced to E and AC is produced to F. BD is bisector of angle CBE and CD is bisector of angle BCF. BD and CD meet at D.

In ABC,

AB = AC[Given]

C = B[angles opposite to equal sides are equal]

CBE = 1800 - B[ABE is a straight line]

[BD is bisector of CBE]

Similarly,

BCF = 1800 - C[ACF is a straight line]

[CD is bisector of BCF]

Now,

In BCD,

BD = CD

In ABD and ACD,

AB = AC[Given]

AD = AD[Common]

BD = CD[Proved]

Therefore, AD bisectsA.

Solution 19

In ABC,

CX is the angle bisector of C

ACY = BCX ....... (i)

In AXY,

AX = AY [Given]

AXY = AYX …….(ii) [angles opposite to equal sides are equal]

Now XYC = AXB = 180° [straight line]

AYX + AYC = AXY + BXY

AYC = BXY ........ (iii) [From (ii)]

In AYC and BXC

AYC + ACY + CAY = BXC + BCX + XBC = 180°

CAY = XBC [From (i) and (iii)]

CAY = ABC

Solution 20

Since IA || CP and CA is a transversal

CAI = PCA [Alternate angles]

Also, IA || CP and AP is a transversal

IAB = APC [Corresponding angles]

But CAI = IAB [Given]

PCA = APC

AC = AP

Similarly,

BC = BQ

Now,

PQ = AP + AB + BQ

= AC + AB + BC

= Perimeter of ABC

Solution 21

In ABD,

BAE = 3 + ADB

1080 = 3 + ADB

But AB = AC

3 = 2

1080 = 2 + ADB ……(i)

Now,

In ACD,

2=1+ ADB

But AC = CD

1 = ADB

2 = ADB + ADB

2 = 2ADB

Putting this value in (i)

1080 = 2ADB + ADB

3ADB = 1080

ADB = 360

Solution 22

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 23

In right BEC and BFC,

BE = CF[Given]

BC = BC[Common]

BEC = BFC[each = 900]

therefore increment BEC space approximately equal to increment BFC space left square bracket RHS right square bracket
rightwards double arrow angle straight B equals angle straight C
Similarly comma
angle straight A equals angle straight B
Hence comma space angle straight A equals angle straight B equals angle straight C
rightwards double arrow AB equals BC equals AC

Therefore, ABC is an equilateral triangle.

Solution 24

DA || CE[Given]

[Corresponding angles]

[Alternate angles]

But [ AD is the bisector of A]

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

AC = AE

ACE is an isosceles triangle.

Solution 25

Produce AD upto E such that AD = DE.

Hence, ABC is an isosceles triangle.

Solution 26

Since AB = AD = BD

is an equilateral triangle.

Again in

AD = DC

Solution 27

(i)

 

 

  

 

 

(ii)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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