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