Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

Ex. 10(A)

Ex. 10(B)

Isosceles Triangle Exercise Ex. 10(A)

Solution 1

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 360^o.

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 = 90^o[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, ΔBPC $approximately equal to$ ΔCQB [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]

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