SELINA Solution Class 9 Chapter 10 Isosceles Triangles Exercise 10B

Question 1

If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.

Sol:

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

In ΔABC,
AB = AC ........[ Given ] 
∴ ∠C = ∠B .....(i) [angels opp. to equal sides are equal]

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

∠ABC + ∠DBC =180° .......[ABD is a st. line] 
∠DBC = 180° − ∠ABC
∠DBC = 180° − ∠B ......(ii)

Similarly,
∠ACB + ∠ECB = 180° .......[ABD is a st. line] 
∠ECB = 180° − ∠ACB
∠ECB = 180° − ∠C ........(iii)
∠ECB = 180° − ∠B .......(iv) [from(i) and (iii)]
∠DBC = ∠ECB ........[from (ii) and (iv)]

Now,
∠DBC = 180° − ∠B 
But ∠B = Acute angel
∴  ∠DBC = 180° − Acute angle = obtuse angle

Similarly,
∠ECB = 180° − ∠C.
But ∠C = Acute angel
∴  ∠ECB = 180° − Acute angle = obtuse angle

Therefore, exterior angles formed are obtuse and equal.

Question 2

In the given figure, AB = AC.

Prove that:
(i) DP = DQ
(ii) AP = AQ
(iii) AD bisects angle A

Sol:

Const: Join AD.

In ΔABC,
AB = AC .......[Given]
∴ ∠C = ∠B  ......(i) [angles opp. to equal sides are equal]

(i)
In ΔBPD and ΔCQD,
∠BPD = ∠CQD ......[Each = 90°]
∠B = ∠C ........[proved]
BD = DC ........[Given]
∴ ΔBPD ≅ ΔCQD .......[AAS criterion]
∴ DP = DQ ......[c.p.c.t]

(ii) 
We have already proved that ΔBPD ≅ ΔCQD
Therefore, BP = CQ .......[c.p.c.t]
Now,
AB = AC .......[Given]
⇒ AB − BP = AC − CQ
⇒ AP = AQ

(iii)
In  ΔAPD and ΔAQD,
DP = DQ .......[proved]
AD = AD .......[common]
AP = AQ ........[Proved]
∴ ΔAPD ≅ ΔAQD ......[SSS]
⇒ ∠PAD = ∠QAD .......[c.p.c.t]

Hence, AD bisects angle A.

Question 3

In triangle ABC, AB = AC; BE ⊥ AC and CF ⊥ AB.

Prove that:
(i) BE = CF
(ii) AF = AE

Sol:

(i)
In ΔAEB and ΔAFC,
∠A = ∠A .......[Common]
∠AEB = ∠AFC = 90° ......[Given: BE ⊥ AC]
                                 ......[Given: CF ⊥ AB]
AB = AC .......[Given]
⇒ ΔAEB ≅ AFC .......[AAS]
∴ BE = CF .......[C.p.c.t]

(ii) Since  ΔAEB ≅ AFC
∠ABE = ∠AFC
∴ AF = AE ........[congruent angles of congruent triangles]

Question 4

In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD.
Prove that: ∠BCD = 90°

Sol:

Const: Join CD.
In ΔABC,
AB = AC .........[ Given ]
∴  ∠C = ∠B .......(i) [angles opp. to equal sides are equal]

In ΔACD,
AC= AD ...[Given]
∴  ∠ADC = ∠ACD ........(ii)

Adding (i) and (ii)
∠B + ∠ADC = ∠C + ACD
∠B + ∠ADC = ∠BCD ....(iii)

In ΔBCD,
∠B + ∠ADC + ∠BCD = 180°
∠BCD + ∠BCD = 180° .......[From (iii)]
2∠BCD = 180°
∠BCD = 90°

Question 5.1

In triangle ABC, AB = AC and ∠A= 36°. If the internal bisector of ∠C meets AB at point D, prove that AD = BC.

Sol:

AB = AC
ΔABC is an isosceles triangle.
∠A = 36°
∠B = C = ${\displaystyle \frac{180°-36°}{2}}$ = 72°

∠ACD = ∠BCD = 36° .......[∵ CD is the angle bisector of ∠C]
ΔADC is an isoscelsss traingle since ∠DAC = ∠DCA = 36°
∴  AD = CD .......(i)

In ΔDCB,
∠CDB = 180° − ( ∠DCB +∠DBC )
        = 180° − ( 36° + 72° )
        =  180° − 108°
         = 72°

ΔDCB is an isosceles triangle since ∠CDB = ∠CBD = 72°
∴  DC = BC ......(ii)
From (i) and (ii), we get
AD = BC
Hence proved.

Question 5.2

If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.

Sol:

Produce AD up to E such that AD = DE.
In ΔABC and ΔEDC,
AD =DE ........[by construction]
BD = CD ...........[Given]
∠1 = ∠2 ..........[verticaly opposite angles]

∴ ΔABD ≅ ΔEDC ......[SAS]
⇒ AB = CE ........(i)
and ∠BAD = ∠CED
But, ∠BAD = ∠CAD .......[AD is bisector of ∠BAC]
∴ ∠CED = ∠CAD
⇒ AC = CE ........(ii)
From (i) and (ii)
AB = AC
Hence, ABC is an isosceles triangle.

Question 6

Prove that the bisectors of the base angles of an isosceles triangle are equal.

Sol:

In ΔABC,
AB = AC                 ....[ Given ]
∴ ∠C = ∠B             ......(i) [ Angles opp. to equal sides are equal]

⇒ ${\displaystyle \frac{1}{2}\angle \text{C}=\frac{1}{2}}$ ∠B

⇒ ∠BCF = ∠CBE       ...(ii)

In ΔBCE and ΔCBF,
∠C = ∠B                 ...[ From (i) ]
∠BCF = ∠CBE         ...[ From (ii) ]
BC = BC                 ...[ Common ]
∴ ΔBCE ≅ ΔCBF     ...[ AAS ]
⇒ BE = CF              ...[ c. p.c.t ]

Question 7

In the given figure, AB = AC and ∠DBC = ∠ECB = 90°

Prove that: 
(i) BD = CE 
(ii) AD = AE

Sol:

In ΔABC,
AB = AC ......[Given]
∴ ∠ACB = ∠ABC ......[angles opp. to equal sides are equal]
⇒ ∠ABC = ∠ACB .....(i)

∠DBC = ∠ECB = 90° ......[Given]

⇒ ∠DBC = ∠ECB ….(ii)

Subtracting (i) from (ii)
∠DCB − ∠ABC = ∠ECB − ∠ACB
⇒ ∠DBA = ∠ECA ........(iii)

In ΔDBA and ΔECA,
∠DBA = ∠ECA ......[From (iii)]
∠DAB = ∠EAC .......[Vertically opposite angles]
AB = AC ......[Given]
∴ ΔDBA ≅ ΔECA .......[ASA]
⇒ BD = CE ...[c. p. c. t]
Also,
AD = AE ...[c. p. c. t]

Question 8

ABC and DBC are two isosceles triangles on the same side of BC. Prove that:

(i) DA (or AD) produced bisects BC at right angle.

(ii) BDA = CDA.

Sol:

DA is produced to meet BC in L.

In ΔABC,
AB = AC   ...[ Given ]
∴ ∠ACB = ∠ABC.......( i )  ...[ angles opposite to equal sides are equal ]

In ΔDBC, 
DB = DC   ...[ GIven ]
∴ ∠DCB = ∠DBC......( ii )  ...[ angles opposite to equal sides are equal ]

Subtracting (i) from (ii)

∠DCB - ∠ACB = ∠DBC - ∠ABC
⇒ ∠DCA = ∠DBA......( iii )

In ΔDBA and ΔDCA,
DB = DC  ...[ GIven ]
∠DBA = ∠DCA ... [ From ( iii ) ]
AB = AC ...[ Given ]
∴ ΔDBA≅ΔDCA ...[ SAS]
⇒ ∠BDA = ∠CDA.........( iv ) ...[ c. p. c .t ]

In ΔDBA, 
∠BAL = ∠DBA + ∠BDA.......( v ) ...[ Ext. angle = sum opp. int. angles]

From (iii), (iv) and (v)
∠BAL = ∠DCA + ∠CDA.....( v i )

In ΔDCA,, 
∠CAL = ∠DCA + ∠CDA.......( vii ) ...[ Ext. angle = sum opp. int. angles]

From (vi) and (vii)

∠BAL = ∠CAL.......( viii )

In ΔBAL = ΔCAL,
∠BAL = ∠CAL  ...[FROm ( viii ) ] 
∠ABL = ∠ACL   ...[ From ( i ) ]
AB = AC    ...[ Given ]
∴ ΔBAL ΔCAL ...[ ASA]
⇒ ∠ALB = ∠ALC ...{ c. p . c. t ]
and BL = LC........( i x )  ...[ c. p . c .t ]

Now,

 ∠ALB + ∠ALC = 180°
⇒ ∠ALb +  ∠ALB = 180°
⇒ 2∠ALB =180°
⇒ ∠ALB = 90°
∴ AL ⊥ BC
or DL ⊥ BC and BL = LC
∴ DA produced bisects BC at right angle.

 

Question 9

The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.

Sol:

The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.

The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.

Question 10

Prove that the medians corresponding to equal sides of an isosceles triangle are equal.

SOl:

In ΔABC,
AB = AC .......(Given)
∴ ∠C = ∠B ......(i) [angles opp. to equal sides are equal] 

⇒ ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}}$
⇒  BF = CE .........(ii)

In ΔBCE and ΔCBF,
∠C = ∠B .......[From (i)]
BF = CE .........[From (ii)] 
BC = BC .......[Common]
∴ ΔBCE ≅ ΔCBF .......[SAS] 
⇒ BE = CF .......[c.p.c.t.]

Question 11

Use the given figure to prove that, AB = AC.

Sol:

In ΔAPQ,
AP = AQ .........[Given]
∴ ∠APQ = ∠AQP .........(i) [angles opposite to equal sides are equal]

In ΔABP,
∠APQ = ∠BAP + ∠ABP   .......(ii) [Ext. the angle is equal to the sum of opp. int. angles]

In ΔAQC,
∠AQP = ∠CAQ + ∠ACQ   ...(iii) [Ext. angle is equal to sum of opp. int. angles]
From (i), (ii) and (iii)
∠BAP + ∠ABP = ∠CAQ + ∠ACQ
But, ∠BAP = ∠CAQ .......[Given]
⇒ ∠CAQ + ∠ABP = ∠CAQ + ∠ACQ
⇒ ∠ABP = ∠CAQ + ∠ACQ − ∠CAQ
⇒ ∠ABP = ∠ACQ
⇒ ∠B = ∠C ...........(iv)

In ΔABC,
∠B = ∠C
⇒ AB = AC ......[Sides opposite to equal angles are equal]

Question 12

Use the given figure to prove that, AB = AC.

SOl:

In ΔAPQ,
AP = AQ .........[Given]
∴ ∠APQ = ∠AQP .........(i) [angles opposite to equal sides are equal]

In ΔABP,
∠APQ = ∠BAP + ∠ABP   .......(ii) [Ext. the angle is equal to the sum of opp. int. angles]

In ΔAQC,
∠AQP = ∠CAQ + ∠ACQ   ...(iii) [Ext. angle is equal to sum of opp. int. angles]
From (i), (ii) and (iii)
∠BAP + ∠ABP = ∠CAQ + ∠ACQ
But, ∠BAP = ∠CAQ .......[Given]
⇒ ∠CAQ + ∠ABP = ∠CAQ + ∠ACQ
⇒ ∠ABP = ∠CAQ + ∠ACQ − ∠CAQ
⇒ ∠ABP = ∠ACQ
⇒ ∠B = ∠C ...........(iv)

In ΔABC,
∠B = ∠C
⇒ AB = AC ......[Sides opposite to equal angles are equal]

Question 12

In the given figure; AE bisects exterior angle CAD and AE is parallel to BC.

Prove that: AB = AC.

SOl:

Since AE || BC and DAB is the transversal.
∴ ∠DAE = ∠ABC = ∠B .......[Corresponding angles]
Since AE || BC and AC are the transversals.
∴ ∠CAE = ∠ACB = ∠C ......[Alternate angles]
But AE bisects ∠CAD,
∴ ∠DAE = ∠CAE
⇒ ∠B = ∠C
⇒  AB = AC .......[Sides opposite to equal angles are equal]

Question 13

In an equilateral triangle ABC; points P, Q and R are taken on the sides AB, BC and CA respectively such that AP = BQ = CR. Prove that triangle PQR is equilateral.

SOl:

AB = BC = CA .......(i) [Given]
AP = BQ = CR .......(ii) [Given]
Subtracting (ii) from (i)
AB − AP = BC − BQ = CA − CR
BP = CQ = AR .......…(iii)
∴ ∠A = ∠B = ∠C .......(iv) [angles opp. to equal sides are equal]

In ΔBPQ and ΔCQR,
BP = CQ ........[From (iii)]
∠B = ∠C .....[From (iv)]
BQ = CR .......[Given]
∴ ΔBPQ ≅ ΔCQR .......[SAS criterion] 
⇒ PQ = QR ........(v)

In ΔCQR and ΔAPR,
CQ = AR .......[From (iii)]
∠C = ∠A ......[From (iv)]
CR = AP .......[Given]
∴ ΔCQR ≅ ΔAPR ...[SAS criterion] 
⇒ QR = PR ...(vi)

From (v) and (vi)
PQ = QR = PR
Therefore, PQR is an equilateral triangle.

Question 14

In triangle ABC, altitudes BE and CF are equal. Prove that the triangle is isosceles.

Sol:

In ΔABE and ΔACF,
∠A = ∠A .........[Common]
∠AEB = ∠AFC = 90° ......[Given: BE ⊥ AC; CF ⊥ AB]
BE = CF ..........[Given]

∴ ΔABE ≅ ΔACE .........[AAS Criterion]
⇒ AB = AC
Therefore, ABC is an isosceles triangle.

Question 15

Through any point in the bisector of an angle, a straight line is drawn parallel to either arm of the angle. Prove that the triangle so formed is isosceles.

Sol:

AL is the 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 = ∠DA ........(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.

Question 16.1

In triangle ABC; AB = AC. P, Q, and R are mid-points of sides AB, AC, and BC respectively.
Prove that: PR = QR

SOl:

In ΔABC,
AB = AC
⇒ ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}}$
⇒ AP = AQ                     ...(i)[ Since P and Q are mid - points ]

In ΔBCA,
PR = ${\displaystyle \frac{1}{2}\text{AC}}$                 ...[ PR is line joining the mid - points of AB and BC ]
⇒ PR = AQ                      ...(ii)

In ΔCAB,
QR = ${\displaystyle \frac{1}{2}\text{AB}}$                 ...[ QR is line joining the mid - points of AC and BC ]
⇒ QR = AP                       …(iii)
From (i), (ii) and (iii)
PR = QR.

Question 16.2

In triangle ABC; AB = AC. P, Q, and R are mid-points of sides AB, AC, and BC respectively.
Prove that: BQ = CP

Sol:

AB = AC
⇒ ∠B = ∠C
Also,
${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}}$
⇒ BP = CQ                  ...[ P and Q are mid-points of AB and AC ]

In ΔBPC and ΔCQB,
BP = CQ
∠B = ∠C
BC = BC
Therefore, ΔBPC ≅ ΔCQB    ...[ SAS ]
BP = CP.

Question 17.1

From the following figure,

prove that: ∠ACD = ∠CBE

Sol:

In ΔACB,
AC = AC ....[Given]
∴ ∠ABC = ∠ACB  ...(i) [angles opposite to equal sides are equal]
∠ACD + ∠ACB = 180° ......(ii) [DCB is a straight line]
∠ABC + ∠CBE = 180°  ….(iii)[ ABE is a straight line ]
Equating (ii) and (iii)
∠ ACD + ∠ACB = ∠ABC + ∠CBE
⇒ ∠ACD + ∠ACB = ∠ACB + ∠CBE ......[From (i)]
⇒ ∠ACD = ∠CBE.

Question 17.2

From the following figure,

prove that: AD = CE.

SOl:

In ΔACD and ΔCBE,
DC= CB .......[Given]
AC = BE .......[Given]
∠ACD = ∠CBE .......[Proved Earlier]
∴ ΔACD ≅ ΔCBE .......[SAS Criterion]
⇒ AD= CE ......[ c.p.c.t. ]

Question 18

Equal sides AB and AC of an isosceles triangle ABC are produced. The bisectors of the exterior angle so formed meet at D. Prove that AD bisects angle A.

SOl:

AB is produced to E and AC is produced to F. BD is the bisector of angle CBE and CD is the bisector of angle BCF. BD and CD meet at D.
In ΔABC,
AB = AC ........[Given]
∴ ∠C = ∠B .........[angles opposite to equal sides are equal]
∠CBE = 180° − ∠B  .......[ABE is a straight line]

⇒ ∠CBE = ${\displaystyle \frac{180°-\angle \text{B}}{2}}$ ........[BD is bisector of ∠CBE]
⇒ ∠CBE = 90° − ${\displaystyle \frac{\angle \text{B}}{2}}$ .........(i)

Similarly,
∠BCF = 180° − ∠C .......[ACF is a straight line]
⇒  ∠BCD = ${\displaystyle \frac{180°-\angle \text{C}}{2}}$ .......[CD is bisector of ∠BCF]
⇒  ∠BCD = 90° − ${\displaystyle \frac{\angle \text{C}}{2}}$ ........(ii)

Now,
⇒ ∠CBD = 90° −  ${\displaystyle \frac{\angle \text{C}}{2}}$ .......[∵ ∠B = ∠C]
⇒ ∠CBD = ∠BCD

In ΔBCD,
∠CBD = ∠BCD
∴ BD = CD

In ΔABD and ΔACD,
AB = AC ........[Given]
AD = AD ........[Common]
BD = CD ........[Proved]
∴ ΔABD ≅ ΔACD ......[SSS Criterion]
⇒  ∠BAD = ∠CAD ......[c.p.c.t.]
Therefore, AD bisects ∠A.

Question 19

ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY.
Prove that: ∠CAY = ∠ABC.

SOl:

In ABC,
CX is the angle bisector of ∠C
⇒ ∠ACY = ∠BCX .........(i)

In ΔAXY,
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 ΔAYC and ΔBXC
∠AYC + ∠ACY + ∠CAY = ∠BXC + ∠BCX + ∠XBC = 180°
⇒ ∠CAY = ∠XBC .......[From (i) and (iii)]
⇒ ∠CAY = ∠ABC

Question 20

In the following figure; IA and IB are bisectors of angles CAB and CBA respectively. CP is parallel to IA and CQ is parallel to IB.

Prove that:
PQ = The perimeter of the ΔABC.

Sol:

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
PQ = AC + AB + BC
PQ = Perimeter of ΔABC.

Question 21

Sides AB and AC of a triangle ABC are equal. BC is produced through C up to a point D such that AC = CD. D and A are joined and produced (through vertex A) up to point E. If angle BAE = 108°; find angle ADB.

SOl:

In ΔABD,
∠BAE = ∠3 + ∠ADB
⇒ 108° = ∠3 + ∠ADB
But AB = AC
⇒ ∠3 = ∠2
⇒ 108° = ∠2 + ∠ADB ....……(i)

Now,
In ΔACD,
∠2 = ∠1 + ∠ADB
But AC = CD
⇒ ∠1 = ∠ADB
⇒ ∠2 = ∠ADB + ∠ADB
⇒ ∠2 = 2∠ADB

Putting this value in (i)
⇒ 108°  = 2∠ADB + ∠ADB
⇒ 3∠ADB = 108° 
⇒ ∠ADB = 36°

Question 22

The given figure shows an equilateral triangle ABC with each side 15 cm. Also, DE || BC, DF || AC, and EG || AB.
If DE + DF + EG = 20 cm, find FG.

Sol:

ABC is an equilateral triangle.
Therefore, AB = BC = AC = 15 cm
∠A = ∠B = ∠C = 60°

In ΔADE, DE || BC ........[ Given ]
∠AED = 60° ........[∵ ∠ACB = 60°]
∠ADE = 60° ........[∵ ∠ACB = 60°]
∠DAE = 180° − (60° + 60°) = 60°
Similarly, BDF and GEC are equilateral triangles.
= 60° .......[∵∠C = 60°]

Let AD = x, AE = x, DE = x ......[∵ ΔADE is an equilateral triangle]
Let BD = y, FD = y, FB = y ......[∵ ΔBDF is an equilateral triangle]
Let EC = z, GC = z , GE = z   ...[∵ΔGEC is an equilateral triangle]

Now,
AD + DB = 15 ⇒  x + y = 15 .......(i)
AE + EC = 15 ⇒ x + z = 15 ........(ii)

Given, DE + DF + EG = 20
⇒ x + y + z = 20
⇒  15 + z = 20 ......[From(i)]
⇒  z = 5
From (ii), we get x = 10
∴ y = 5
Also, BC = 15
BF + FG + GC = 15
⇒ y + FG + z = 15
⇒ 5 + FG + 5 = 15
⇒ FG = 5

Question 23

If all the three altitudes of a triangle are equal, the triangle is equilateral. Prove it.

Sol:

In right ΔBEC and ΔBFC,
BE = CF ........[Given]
BC = BC ........[Common]
∠BEC = ∠BFC ........[each = 90°]
∴ ΔBEC ≅ ∆CFB .........[RHS]
⇒ ∠B = ∠C
Similarly,
∠A = ∠B
Hence, ∠A = ∠B = ∠C
⇒ AB = BC = AC
Therefore, ABC is an equilateral triangle.

Question 24

In a ΔABC, the internal bisector of angle A meets the opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that ΔACE is isosceles.

Sol:

DA || CE                 ... [Given]
⇒ ∠1 = ∠4            ...(i) ( Corresponding angles )
∠2 = ∠3                ....(ii) ( Alternate angles )
But ∠1 = ∠2          ....(iii) ( AD is the bisector of ∠A )

From (i), (ii) and (iii)
∠3 = ∠4
⇒ AC = AE
⇒ ΔACE is an isosceles triangle.

Question 25

In triangle ABC, the bisector of angle BAC meets the opposite side BC at point D. If BD = CD, prove that ΔABC is isosceles.

SOl:

Produce AD up to E such that AD = DE.
In ΔABD and ΔEDC,
AD = DE                     ...[ by construction ]
BD = CD                     ...[ Given ]
∠1= ∠2                       ...[ Vertically opposite angles ]
∴ ΔABD ≅ ΔEDC        ...[ SAS ]
⇒ AB = CE                 ...(i)
and ∠BAD = ∠CED
but, ∠BAD = ∠CAD     ...[ AD is bisector of ∠BAC ]
∴ ∠CED = ∠CAD
⇒ AC = CE                  ...(ii)
From (i) and (ii)
AB = AC
Hence, ABC is an isosceles triangle.

Question 26

In ΔABC, D is point on BC such that AB = AD = BD = DC.
Show that: ∠ADC : ∠C = 4 : 1.

Sol:

Since, AB = AD = BD
∴ ΔABD is an equilateral triangle.
∴ ∠ADB = 60°
⇒ ∠ADC = 180° − ∠ADB
               = 180° − 60°
               = 120°
Again in ΔADC,
AD = DC
∴ ∠1 = ∠2
But,
∠1 + ∠2 + ∠ADC = 180°
⇒ 2∠1 + 120° = 180°
⇒ 2∠1 = 60°
⇒  ∠1 = 30°
⇒  ∠C = 30°
∴ ∠ADC : ∠C = 120° : 30°
⇒  ∠ADC : ∠C = 4 : 1

Question 27.1

Using the information given of the following figure, find the values of a and b. [Given: CE = AC] 

SOl:

In ΔCAE,
∠CAE = ∠AEC = ${\displaystyle \frac{180°-68°}{2}}$ = 56°    [∵ CE = AC] 

In ∠BEA,  a = 180° − 56° = 124°

In ∠ABE, ∠ABE = 180° − (a + ∠BAE)
                         = 180° − (124° + 14°)
                        = 180° − 138° = 42°

Question 27.2

Using the information given of the following figure, find the values of a and b.

Sol:

In ΔAEB and ΔCAD,
∠EAD = ∠CAD .........[Given]
∠ADC = ∠AEB .......[∵ ∠ADE = ∠AED { AE = AD }180° − ∠ADE = 180° − ∠AED = ∠ADC = ∠AEB] 
AE = AD .........[Given]
∴ ΔAEB ≅ ΔCAD ....[ASA]
AC = AB .......[By C.P.C.T.]
2a + 2 = 7b − 1
⇒ 2a − 7b = − 3 ....(i)
CD = EB
⇒ a = 3b ....(ii)
Solving (i) and (ii), We get,
a = 9, b = 3

SELINA Solution Class 9 Chapter 10 Isosceles Triangles Exercise 10A

Question 1

In the figure alongside,

AB = AC
∠A = 48°  and
∠ACD = 18° 
Show that BC = CD.

Sol:

In ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
48° + ∠ACB + ∠ABC = 180°
But ∠ACB = ∠ABC ........[ AB = AC ]
2∠ABC = 180° - 48°
2∠ABC = 132°
∠ABC = 66° = ∠ACB ..........(i)

∠ACB = 66°
∠ACD + ∠DCB = 66°
18° + ∠DCB = 66°
∠DCB = 48° ..........(ii)

Now, In ΔDCB,
∠DBC = 66° .......[ From (i), Since ∠ABC = ∠DBC ]
∠DCB = 48° .......[From (ii)]
∠BDC = 180°- 48° - 66°
∠BDC = 66°
Since ∠BDC = ∠DBC
Therefore, BC = CD
Equal angles have equal sides opposite to them.

Question 2

Calculate:
(i) ∠ADC
(ii) ∠ABC
(iii) ∠BAC

Sol:

Given: ACE = 130°; AD = BD = CD

Proof:
(i) ∠ACD + ∠ACE = 18° ....... [ DCE is a st. line ]
⇒ ∠ACD = 180° − 130°
⇒ ∠ACD = 50° 
Now, CD = AD
⇒ ∠ACD = ∠DAC = 50°  ..... (i)[ Since angels opposite to equal sides are equal]

In ΔADC,
∠ACD = ∠DAC = 50°  
∠ACD + ∠DAC + ∠ADC = 180° 
 50°  + 50°  + ∠ADC = 180° 
∠ADC =180° − 100°
∠ADC = 80°

(ii) ∠ADC = ∠ABD + ∠DAB ....[Exterior angle is equal to sum of opp. interor angle] 
But AD = BD
∴ ∠DAB = ∠ABD
⇒ 80° = ∠ABD + ∠ABD
⇒ 2∠BD = 80°
⇒ ∠ABD = 40° = ∠DAB .....(ii)

(iii) ∠BAC = ∠DAB + ∠DAC
substituting the value from (i) and (ii)
∠BAC = 40° + 50°
⇒ ∠BAC = 90°

Question 3

In the following figure, AB = AC; BC = CD and DE are parallel to BC.
Calculate:
(i) ∠CDE
(ii) ∠DCE

Sol:

∠FAB = 128° .......[ Given ]
∠BAC + ∠FAB = 180° ......[ FAC is a st. line ] 
⇒ ∠BAC = 180° − 128° 
⇒ ∠BAC = 52° 

In ΔABC,
∠A = 52°
∠B=  ∠C .....[Given AB = AC and angels opposite to equal sides are equal]

∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + ∠B = 180° 
⇒ 52° + 2∠B = 180°
⇒ 2∠B = 128°
⇒ ∠B = 64° = ∠C ........(i)
⇒ ∠B = ∠ADE .......[ Given DE || BC ]
(i) 
Now,
∠ADE + ∠CDE + ∠B = 180° ....[ ADB is a st. line ]
⇒  64° + ∠CDE + 64° = 180°
⇒ ∠CDE = 180° − 128° 
⇒ ∠CDE = 52° 

(ii)
Given DE || BC and DC is the transversal.
⇒ ∠CDE = ∠DCB = 52° .......(ii)
Also, ∠ECB = 64° .....[ From (i) ]
But,
∠ECB = ∠DCE + ∠DCB
⇒ 64° = ∠DCE + 52°
⇒ ∠DCE + 64° − 52°
⇒  ∠DCE = 12°

Question 4.1

Calculate x :

Sol:

Let the triangle be ABC and the altitude be AD.

In ΔABD,
∠DBA = ∠DAB = 37° .....[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB .......[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 37° + 37° 
⇒ ∠CDA = 74°
Now in ΔADC,
∠CDA = ∠CAD = 74° ....[ Given CD = AC and angles opposite to equal sides are equal]
Now,
∠CAD + ∠CDA + ∠ACD = 180°
⇒ 74° + 74° + x = 180° 
⇒ x = 180° − 148° 
⇒ x = 32°

Question 4.2

Calculate x :

Sol:

Let triangle be ABC and altitude be AD.

In ΔABD,
∠DBA = ∠DAB = 50° ...[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB ...[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 50° + 50° 
⇒ ∠CDA = 100°

In ΔADC,
∠DAC = ∠DCA = x  ...[Given AD = DC and angles opposite to equal sides are equal]
∴ ∠DAC + ∠DCA + ∠ADC = 180°
⇒ x + x + 100° = 180°
⇒ 2x = 80°
⇒ x = 40°

Question 5

In the figure, given below, AB = AC.
Prove that: ∠BOC = ∠ACD.

Sol:

Let  ∠ABO = ∠OBC = x and ∠ACO = ∠OCB = y
In ΔABC,
∠BAC = 180° - 2x - 2y          ....(i)
Since, ∠B = ∠C                     ...[AB = AC]
${\displaystyle \frac{1}{2}\text{B}=\frac{1}{2}\text{C}}$
⇒ x = y
Now, 
∠ACD = 2x + ∠BAC        ...[ Exterior angle is equal to sum of opp. interior angles ]
∠ACD = 2x + 180° - 2x - 2y      ...[ From(i) ] 
∠ACD =180° - 2y                      ....(i)

In ΔOBC,
∠BOC = 180°- x - y 
⇒ ∠BOC = 180°- y - y             ...[ Already proved ]
⇒ ∠BOC = 180° - 2y                ...(ii)

From (i) and (ii)
∠BOC = ∠ACD

Question 6

In the figure given below, LM = LN; angle PLN = 110^o.

calculate: (i) ∠LMN
                 (ii) ∠MLN

Sol:

Given: ∠PLN = 110°
(i) We know that the sum of the measure of all the angles of a quadrilateral is 360°. 
In quad. PQNL,
∠QPL + ∠PLN + ∠LNQ + ∠NQP = 360°
⇒ 90° + 110° + ∠LNQ + 90° = 360°
⇒ ∠LNQ = 360° − 290°
⇒ ∠LNQ = 70°
⇒ ∠LNM = 70° ........(i)
In ΔLMN,
LM = LN ........( Given )
∴ ∠LNM = ∠LMN ....... [angles opp. to equal sides are equal]
⇒ ∠LMN = 70° ....(ii) [ from(i) ]

(ii) In ΔLMN,
∠LMN + ∠LNM+ ∠MLN = 180°
But ∠LNM= ∠LMN = 70° .....[ From(i) and (ii)]
∴ 70° + 70° + MLN = 180°
⇒ ∠MLN = 180°− 140°
⇒ ∠MLN = 40°

Question 7

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠ CAB = 55^o.
Find:
(i) ∠DCB 
(ii) ∠CBD.

Sol:

ln ΔABC,
AC = BC .......[Given]
∴ ∠CAB = ∠CBD ........[angles opp.to equal sides are equal]
⇒ ∠CBD = 55°

In ΔABC,
∠CBA + ∠CAB + ∠ACB = 180°
but, ∠CAB = ∠CBA = 55°
⇒ 55° + 55° + ∠ACB = 180°
⇒  ∠ACB = 180° − 110°
⇒  ∠ACB = 70°
Now,
In ΔACD and ΔBCD,
AC = BC .......[Given]
CD = CD ........[Common]
AD = BD .........[Given: CD bisects AB]
∴ ΔACD ≅ ΔBCD
⇒ ∠DCA = ∠DCB
⇒ ∠DCB = ${\displaystyle \frac{\angle \text{ACB}}{2}=\frac{70°}{2}}$
⇒ ∠DCB = 35°.

Question 8

Find x :

Sol:

Let us name the figure as following :

In ΔABC, 
AD = AC .......[ Given ] 
∴ ∠ADC =∠ACD ......[Angles opp. to equal sides are equal]
⇒ ∠ADC = 42°

Now, 
∠ADC = ∠DAB +∠DBA ...[Exterior angle is equal to the sum of opp. interior angles]
But, 
∠DAB = ∠DBA ......[Given: BD = DA]
∴ ∠ADC = 2∠DBA
⇒ 2∠DBA = 42°
⇒ ∠DBA = 21°

For x:
x = ∠CBA + ∠BCA .......[Exterior angle is equal to the sum of opp. interior angles]
We know that,
∠CBA = 21°
∠BCA = 42°
∴ x = 21 + 42°
⇒ x = 63°

Question 9

In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.

Sol:

In ΔABD aand ΔDBC,
BD  = BD                 ...[ Common ]
∠BDA = ∠BDC        ...[ each equal to 90° ]
∠ABD = ∠DBC        ...[ BD bisects ∠ABC ]
∴ ΔABD ≅ ΔDBC       ...[ ASA criterion ]

Therefore,
AD = DC
x + 1 = y + 2
⇒ x = y + 1               ..... (i)
and AB = BC
3x + 1 = 5y - 2

Subtituting the value of x from (i)
3( y + 1 ) + 1 = 5y - 2 
⇒ 3y + 3 + 1 = 5y - 2
⇒ 3y +  4 = 5y - 2
⇒ 2y = 6
⇒ y = 3

Putting y = 3 in (i)
x = 3 + 1 
∴ x = 4

Question 10

In the given figure; AE || BD, AC || ED and AB = AC. Find ∠a, ∠b and ∠c.

Sol:

Let P and Q be the points as shown below: 

Given:
∠PDQ = 58°
∠PDQ = ∠EDC = 58° .....[ Vertically opp . angles ]
∠EDC = ∠ACB =58° ........[ Corresponding angles  ∵ AC || ED ]

In  ΔABC,
AB = AC .......[ Given ]
∴ ∠ACB = ∠ABC = 58° ......[angels opp. to equal sides are equal]
Now, 
∠ACB + ∠ABC+ ∠BAC = 180°
⇒ 58° + 58° + a = 180°
⇒ a = 180° − 116°
⇒ a = 64°
Since AE || BD and AC is the transversal. 
∠ABC = b .......[ Corresponding angles ]
∴ b = 58°

Also since AE || BD and ED is the transversal 
∠EDC = c  .......[ Corresponding angles ] 
∴ c = 58°

Question 11

In the following figure; AC = CD, AD = BD and ∠C = 58^o.

Find the angle CAB.

SOl:

In ΔACD,
AC = CD                      ...[ Given]
∴ ∠CAD = ∠CDA
∠ACD = 58°                ...[ Given ]

∠ACD + ∠CDA + ∠CAD = 180°
⇒ 58° + 2∠CAD = 180°
⇒ 2∠CAD = 122°
⇒ ∠CAD = ∠CDA = 61° ...(i) 

Now,
∠CDA = ∠DAB + ∠DBA ...[ Ext. angel is equal to sum of opp. int. angles ]
But,
∠DAB = ∠DBA                ...[ Given : AD = DB ]
∴ ∠DAB +∠ DAB = ∠CDA
⇒ 2∠DAB = 61°
⇒ ∠DAB = 30.5°             ....(ii)

In ΔABC,
∠CAB = ∠CAD +∠DAB
∴ ∠CAB = 61° + 30.5°
⇒ ∠AB = 91.5°

Question 12

In the figure given below, if AC = AD = CD = BD; find angle ABC.

Sol:

In ΔACD,
AC = AD = CD ......[Given]
Hence, ACD is an equilateral triangle.
∴ ∠ACD = ∠CDA = ∠CAD = 60°
∠CDA = ∠DAB + ∠ABD ........[Ext angle is equal to the sum of opp. int. angles]
But,
∠DAB = ∠ABD ......[Given: AD = DB]
∴ ∠ABD + ∠ABD = ∠CDA
⇒ 2∠ABD = 60°
⇒ ∠ABD = ∠ABC = 30°

Question 13

In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.

Sol:

Let ∠A = 8x and ∠B = 5x
Given: AB = AC
⇒ ∠B = ∠C = 5x     ...[Angles opp. to equal sides are equal]

Now,
∠A + ∠B + ∠C = 180°
⇒ 8x + 5x + 5x = 180°
⇒ 18x = 180°
⇒ x = 10°

Given that :
∠A = 8x
⇒ ∠A = 8 x 10°
⇒ ∠A = 80°.

Question 14

In triangle ABC; ∠A = 60^o, ∠C = 40^o, and the bisector of angle ABC meets side AC at point P. Show that BP = CP.

SOl:

In ΔABC,
∠A = 60°
∠C = 40°
∴ ∠B = 180° - 60° - 40°
⇒ ∠B = 80°

Now,
BP is the bisector of ∠ABC.

∴ ∠PBC = ${\displaystyle \frac{\text{∠ABC}}{2}}$

⇒ ∠PBC = 40°
In ΔPBC,
∠PBC = ∠PCB = 40°
∴ BP = CP             ....[ Sides opp. to equal angles are equal.]

Question 15

In triangle ABC; angle ABC = 90^o and P is a point on AC such that ∠PBC = ∠PCB.
Show that: PA = PB.

Sol:

Let PBC = PCB = x
In the right angled triangle ABC,
∠ABC = 90°
∠ACB = x
⇒ ∠BAC = 180° - ( 90° + x )
⇒  ∠BAC = ( 90°- x )             ...(i)

and
∠ABP = ∠ABC - ∠PBC
⇒  ∠ABP = 90° - x                   ...(ii)

Therefore in the triangle ABP;
 ∠BAP = ∠ABP
Hence, PA = PB     ...[sides opp. to equal angles are equal]

Question 16

ABC is an equilateral triangle. Its side BC is produced up to point E such that C is mid-point of BE. Calculate the measure of angles ACE and AEC.

SOl:

ΔABC is an equilateral triangle.
⇒ Side AB = Side AC
⇒ ∠ABC = ∠ACB ........[If two sides of a triangle are equal, then angles opposite to them are equal]

Similarly, Side AC = Side BC
⇒ ∠CAB = ∠ABC .......[If two sides of a triangle are equal, then angles opposite to them are equal]

Hence, ∠ABC = ∠CAB = ∠ACB  = y(say)
As the sum of all the angles of the triangle is 180°.
∠ABC + ∠CAB + ∠ACB = 180°
⇒ 3y = 180°
⇒ y = 60°

∠ACB = ∠ACB = ∠ABC = 60°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠CAB + ∠CBA = ∠ACE
⇒  60° + 60° = ∠ACE
⇒  ∠ACE = 120°

Now ΔACE is an isosceles triangle with AC = CF
⇒ ∠EAC = ∠AEC
Sum of all the angles of a triangle is 180°
∠EAC + ∠AEC + ∠ACE = 180°
⇒ 2∠AEC + 120° = 180°
⇒  2∠AEC = 180° − 120°
⇒  ∠AEC = 30°

Question 17

In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.

Sol:

ΔDBC is an isosceles triangle.
As, Side CD = Side DB
⇒ ∠DBC = ∠DCB .......[If two sides of a triangle are equal, then angles opposite to them are equal]

And ∠B = ∠DBC = ∠DCB = 28°
As the sum of all the angles of the triangle is 180°
∠DCB + ∠DBC + ∠BCD = 180°
⇒ 28° + 28° + ∠BCD = 180°
⇒ ∠BCD = 180° − 56°
⇒ ∠BCD = 124°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠DBC+ ∠DCB = ∠DAC
⇒ 28° + 28° = 56°
⇒ DAC = 56°
Now ΔACD is an isosceles triangle with AC = DC
⇒ ∠ADC = ∠DAC = 56°
Sum of all the angles of a triangle is 180°
⇒ ∠ADC + ∠DAC + ∠DCA = 180°
⇒ 56° +56° + ∠DCA = 180°
⇒ ∠DAC = 180° − 112°
⇒ ∠DCA = 64° = ∠ACD.

Question 18

In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.

SOl:

We can see that the ΔABC is an isosceles triangle with Side AB = Side AC.
⇒ ∠ACB = ∠ABC
As ∠ACB = 65°
hence ∠ABC = 65°
Sum of all the angles of a triangle is 180°
∠ACB + ∠CAB + ∠ABC = 180°
65°+ 65° + ∠CAB = 180°
∠CAB = 180° − 130°
∠CAB = 50°

As BD is parallel to CA
Therefore, ∠CAB = ∠DBA since they are alternate angles.
∠CAB = ∠DBA = 50°
We see that ΔADB is an isosceles triangle with Side AD = Side AB.
⇒ ∠ADB = ∠DBA = 50°
Sum of all the angles of a triangle is 180°
∠ADB + ∠DAB + ∠DBA = 180°
50° + ∠DAB + 50° = 180°
∠DAB = 180° − 100° = 80°
∠DAB = 80°
The angle DAC is the sum of angle DAB and CAB.
∠DAC = ∠CAB + ∠DAB
∠DAC = 50°+ 80°
∠DAC = 130°

Question 19.1

Prove that a triangle ABC is isosceles, if: altitude AD bisects angles BAC.

SOl:

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.

Question 19.2

Prove that a triangle ABC is isosceles, if: bisector of angle BAC is perpendicular to base BC.

Sol:

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 isosceles.

Question 20

In the given figure; AB = BC and AD = EC.
Prove that: BD = BE.

SOl:

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)