SELINA Solution Class 9 Chapter 9 Triangles (Congruency in Triangles) Exercise 9A

Question 1

If the following pair of the triangle is congruent? state the condition of congruency : 
In Δ ABC and Δ DEF, AB = DE, BC = EF and ∠ B = ∠ E.

Sol:

In Δ ABC and Δ DEF,
AB = DE         ...[ Given ]
∠B = ∠E       ...[ Given ]
BC = EF          ...[ Given ]

By Side - Angle - Side criterion of congruency, the triangles
Δ ABC and Δ DEF  are congruent to each other.
∴ Δ ABC ≅ Δ DEF

Question 1.2

If the following pair of the triangle is congruent? state the condition of congruency : 
In  ΔABC and ΔDEF,  ∠B = ∠E = 90^o; AC = DF and BC = EF.

Sol:

In Δ ABC and Δ DEF
∠B = ∠E = 90^ο 
Hyp. AC= HYp. DF
BC = EF
By Right Angel- Hypotenuse-Side Criterion of  congruency, the triangles
Δ ABC and Δ DEF are congruent to each other.
∴ Δ ABC ≅ Δ DEF

Question 1.3

If the following pair of the triangle is congruent? state the condition of congruency: 

In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.

Sol:

In ΔABC and ΔQRP 
∠B =∠R [ Given ] 
∠C =∠P [ Given ] 
AB = QR [ Given ]

By Angel-AAngel SIde criterion of congruency, the triangles 
ΔABC and ΔQRP  are congruent to each other.
∴ ΔABC  ≅ ΔQRP

Question 1.4

If the following pair of the triangle is congruent? state the condition of congruency: 

In ΔABC and ΔPQR, AB = PQ, AB = PQ, and BC = QR.

Sol:

In ΔABC and ΔPQR
AB = PQ [ Given ]
AB = PQ [ Given ]
BC = QR [ Given ]

By Side- Side - Side criterion of congruency, the triangles
ΔABC and ΔPQR are congruent to each other.
∴  ΔABC  ≅ ΔPQR

Question 1.5

If the following pair of the triangle is congruent? state the condition of congruency: 

In ΔADC and ΔPQR, BC = QR, ∠A = 90^o, ∠C = ∠R = 40^o and ∠Q = 50^o.

Sol:

In ΔPQR
∠R = 40^o , ∠Q = 50^o
∠P+∠Q+∠R=180^o       [ Sum of all the angels in aa triangle = 180^o ]

⇒ ∠P+ 50^o+ 40^o = 180^o  
⇒        ∠P + 90^o  = 180^o   
⇒                ∠P = 180^o  -  90^o 
⇒                ∠P = 90^o  
In ΔADC and ΔPQR
∠A = ∠P
∠C = ∠R
BC = QR  

By Angle-Angle- Side criterion of congruency, the triangles
ΔADC and ΔPQR are congruent to each other.
∴ ΔADC ≅ ΔPQR

Question 2

The given figure shows a circle with center O. P is mid-point of chord AB.

Show that OP is perpendicular to AB.

Sol:

Given: in the figure, O is center of the circle, and AB is chord. P is a point on AB such that AP = PB.
We need to prove that, OP ⊥ AB

Construction: Join OA and OB
Proof:  
In ΔOAP and ΔOBP
OA=OB        ...[radii of the same circle]
OP=OP        ...[common]
AP=PB         ...[given]
∴ By Side-Side-Side criterion of congruency,
ΔOAP ≅  ΔOBP 
The corresponding parts of the congruent
triangles are congruent.
∴ ∠OPA=∠OPB                     ...[by c.p.c.t]
But ∠OPA + ∠OPB += 180⁰ ....[linear pair]
∴ ∠OPA= ∠OPB= 90⁰
Hence OP ⊥ AB.

Question 3

The following figure shows a circle with center O.

If OP is perpendicular to AB, prove that AP = BP.

Sol:

Given: In the figure, O is the center of the circle, and AB is a chord. P is a point on AB such that AP=PB.
We need to prove that, AAP=BP

Construction: Join OA and OB
Proof: 
In right triangles ΔOAP and ΔOBP
Hypotenuse OA=OB      .....[ radii of the same circle ]
Side OP= OP                   ...[ common ]
∴ By Right Angle- Hypotenuse- Side criterion of congruency, ΔOAP ≅ ΔOBP
The corresponding parts of the congruent triangles are congruent.
∴ AP=BP                        .....[ by c.p.c.t ]
Hence proved.

Question 4

In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent. 
(ii) AB = CE.
(iii) AB is parallel to EC

Sol:

Given: A ΔABC in which D is the mid-point of BC
AD is produced to E so that DE=AD
WE need to prove that :
(i) ΔABD and ΔECD are congruent. 
(ii) AB = CE.
(iii) AB is parallel to EC

(i) In ΔABD and ΔECD 
BD=DC       ...[ D is the midpoint of BC ]
ADB=CDE   ...[ vertically opposite angels ]
AD=DE       ...[ Given ]

∴ By Side-Angel-Side criterion of congruence, we have,
ΔABD ≅ ΔECD

(ii) The corresponding parts of the congruent triangles are congruent.
∴ AB=EC        ...[ c.p.c.t ] 

(iii) Also, DAB = DEC    ....[ c.p.c t ]

AB || EC  .....[ DAB and DEC are alternate angels ] 

Question 5.1

A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.

Sol:

Given: A ΔABC in which ∠B = ∠C.
DL is the perpendicular from D to AB
DM is the perpendicular from D to AC

We need to prove that
DL = DM
Proof:
In ΔDLB and ΔDMC
∠DLB = ∠DMC=90⁰  ...[ DL ⊥ AB and DM ⊥ AC ]
∠B=∠C                      ...[ Given ]
BD= DC                    ...[ D is the midpoint of BC ]
∴ By Angel-Angel-SIde Criterion of congruence, 
ΔDLB ≅ ΔDMC
The corresponding parts of the congruent triangles are congruent.
∴DL=DM                ...[ c.p.c.t ]

Question 5.2

A triangle ABC has B = C.
Prove that: The perpendicu...lars from B and C to the opposite sides are equal.

Sol:

Given: A ΔABC in which ∠B = ∠C.
BP is perpendicular from D to AC
CQ is the perpendicular from C to AB

We need to prove that
BP = CQ
Proof:
In  ΔBPC and  ΔCQB
∠B= ∠C                   ...[ Given ]
∠BPC= ∠CQB=90  ...[ BP AC and CQ AB ]
BC =BC                  ...[ Common ]
∴ BY Angel-Angel-Side criterion of congruence,
ΔBPC ≅ ΔCQB
The corresponding parts of the congruent triangles are congruent.
BP = CQ                   ...[ c .p.c.t ]

Question 6

The perpendicular bisectors of the sides of a triangle ABC meet at I.

Prove that: IA = IB = IC.

Sol:

Given: A  ΔABC in which AD is the perpendicular bisector of BC 
BE is the perpendicular bisector of CA
CF is the perpendicular bisector of AB
AD, BE and CF meet at I

WE need to prove that
IA = IB= IC
Proof:
In  ΔBID and  ΔCID
BD = DC                 ...[ Given ]
∠BDI = ∠CDI = 90°...[ AD is the perpendicular bisector of BC]
DI = DI               ...[ Common ]
∴ By the Side-Angle-Side criterion of congruence,
 Δ BID ≅ Δ CID
The corresponding parts of the congruent triangles are congruent.
∴ IB = IC               ...[ c.p.c.t ]

Similarly, in  Δ CIE and  Δ AIE
CE = AE              ...[ Given ]
∠CEI = ∠AEI = 90° ...[ AD is the perpendicular bisector of BC ]
IE = IE                ...[ Common ]
∴ By Side-Angel-Side Criterion of congruence,
 ΔCIE ≅ ΔAIE
The corresponding parts of the congruent triangles are congruent.
∴ IC = IA         ...[ c.p.c.t ]

Thus, IA = IB = IC

Question 7

A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.

Sol:

Given: A ΔABC in which AB is bisected at P 
PQ is perpendicular to AB

WE need to prove that
QA = QB
Proof:
In ΔAPQ and ΔBPQ
AP = PB                    ...[ P is the mid-point of AB ]
∠APQ = ∠BPQ = 90°  ...[ PQ is perpendicular to AB ]
PQ = PQ                     ...[ Common] 
∴ By Side-Angel-Side criterion of congruence,
ΔAPQ ≅ ΔBPQ
The corresponding parts of the congruent triangles are congruent.
∴ QA = QB   ...[ c.p.c.t ]

Question 8

If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.

Sol:

From M, draw ML such that ML is perpendicular to AB and MN is perpendicular to AC

In ΔALM and ΔANM
∠LAM = ∠MAN          ...[ ∵ AP is the bisector of BAC ]
∠ALM = ∠ANM = 90°  ...[ ∵ ML ⊥ AB, MN ⊥ AC ]
AM = AM                   ...[ Common ]
∴ By Angel-Angel-Side criterion of congruence, 
ΔALM ≅ ΔANM
The corresponding parts of the congruent triangles are congruent.
∴ ML = MN               ...[ c. p. c. t ]
Hence proved.

Question 9

From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that: 
(i) ΔDCE ≅ ΔLBE 
(ii) AB = BL.
(iii) AL = 2DC

Sol:

Given: ABCD is a parallelogram in which is the mid-point of BC.
We need to prove that
(i) ΔDCE ≅ ΔLBE 
(ii) AB = BL.
(iii) AL = 2DC

(i) In  ΔDCE and ΔLBE
∠DCE = ∠EBL   ...[DC || AB, alternate angels]
CE = EB             ...[ E is the midpoint of BC]
∠DEC= ∠LEB    ...[ vertically opposite angels]
∴ By Angel-SIde-Angel Criterion of congruence, we have,
 ΔDCE ≅ ΔLBE
The corresponding parts of the congruent triangles are congruent.
∴ DC= LB        ...[ c. p. c .t]       ....(1)

(ii) DC= AB  ...[ opposite sides of a parallelogram]...(2)
From ( 1 ) and ( 2 ), Ab = BL                      ...(3)

(iii) Al =  AB+ BL                                      ... (4)    
From (3) and (4), Al = AB + AB
⇒AL = 2AB
⇒AL = 2DC                    ...[ From (2) ]

Question 10

In the given figure, AB = DB and Ac = DC.

If ∠ ABD = 58^o,
∠ DBC = (2x - 4)^o,
∠ ACB = y + 15^o and
∠ DCB = 63^o ; find the values of x and y.

Sol:

Given:
In the figure AB = DB, AC = DC, ∠ABD =  58°,
∠DBC = ( 2x - 4 )°, ∠ACB = ( y +15)° and ∠DCB = 63°
We need to find the values of x and y. 

In ΔABC and ΔDBC
AB = DB              ...[ Given ]
AC= DC               ...[ Given ]
BC= BC                ...[ common ]
∴ By Side-SIde-Side criterion of congruence, we have,
ΔABC ≅ ΔDBC
The corresponding parts of the congruent triangles are congruent.
∴ ∠ABC= DCB       ...[ c. p. c .t ]
⇒ y° + 15° = 63° 
⇒ y° = 63° - 15°  
⇒ y° = 48°  
and ∠ABC =∠DBC  ...[ c.p.c.t ]
But, ∠DBC = ( 2x  - 4)°  
We have ∠ABC + ∠DBC = ∠ABD
⇒ (2x  - 4)° + (2x - 4)°  = 58°  
⇒  4x - 8^°= 58^°
⇒  4x = 58^° + 8^°
⇒  4x = 66^°
⇒  X = ` 66<sup>°</sup>/(4)`
⇒  X = 16.5^°
Thus the values of x and y are :
x = 16.5°  and y = 48^°

Question 11

In the given figure: AB//FD, AC//GE and BD = CE;
prove that:  (i) BG = DF       (ii) CF = EG    

Sol:

In the given figure AB || FD,
⇒ ∠ABC =∠FDC
Also AC || GE,
⇒ ∠ACB = ∠GEB

Consider the two triangles ΔGBE and ΔFDC
∠B = ∠D
∠C = ∠E
Also given that
BD = CE
⇒ BD + DE = CE + DE
⇒ BE = DC
∴  By Angel-Side-SIde-Angel criterion of congruence
ΔGBE ≅ ΔFDC

∴  ${\displaystyle \frac{\text{GB}}{\text{FD}}=\frac{\text{BE}}{\text{DC}}=\frac{\text{GE}}{\text{FC}}}$
But BE =DC

⇒ ${\displaystyle \frac{\text{BE}}{\text{DC}}=\frac{\text{BE}}{\text{BE}}}$ = 1

∴ ${\displaystyle \frac{\text{GB}}{\text{FD}}=\frac{\text{BE}}{\text{DC}}}$ = 1
⇒  GB = FD

∴ ${\displaystyle \frac{\text{GE}}{\text{FC}}=\frac{\text{BE}}{\text{DC}}}$ = 1
⇒  GE = FC

Question 12

In ∆ABC, AB = AC. Show that the altitude AD is median also.

Sol:

In ∆ABC and ∆ADC,
AB =AC               ....( Since is ani isosceles triangle)
AD =AD              .....( common side )
∠ADB = ∠ADC    ....( Since AD is the altitude so each is 90⁰ )
⇒ΔADB≅ΔADC   .....( RHS congruence criterion ) 
BD = DC              ....( cpct )
⇒ AD is the median.

Question 13

In the following figure, BL = CM.

Prove that AD is a median of triangle ABC.

Sol:

In  ΔDLB and  ΔDMC,
BL = CM                     ...( given )
∠DLB = ∠DMC           ...( Both are 90° )
∠BDL = ∠CDM           ....( vertically opposite angels )
∴ ΔDLB ≅ ΔDMC       ....( AAS congruence criterion )
BD = CD                      ....( cpct )
Hence, AD is the median of ΔABC.

Question 14.1

In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that:  BD = CD

Sol:

In ΔADB and ΔADC,
∠ADB = ∠ADC        ....(Since AD is perpendicular to BC)
AB = AC                  ....(given)
AD = AD                  ....(common side)
 ∴ ΔADB ≅ ΔADC    ....(RHS congruence criterion)
⇒ BD = CD              ....(cpct)

Question 14.2

In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that :  ED = EF 

Sol:

In ΔEFB and ΔEDB,
∠EFB = ∠EDB ( both are 90⁰ )
EB = EB ( common side )
∠FBE = ∠DBE ( given )
ΔEFB ≅ ΔEDB  (AAS congruence criterion)
⇒ EF = ED (cpct )
that is , Ed = EF.

Question 15

Use the information in the given figure to prove :
(i) AB = FE
(ii) BD = CF 

Sol:

ln Δ ABC and Δ EFD,
AB II EF ⇒ ∠ ABC = ∠ EFD  ...( alternate angles )
AC = ED                              ...( given )
∠ ACB = ∠ EDF                    ...( given )
∴ Δ ABC ≅  Δ EFD               ....( AAS congruence criterion )
⇒ AB = FE                           ...( cpct )
and BC= DF                        ....( cpct )
⇒ BD + DC = CF + DC       .....( B-D-C-F )
⇒ BD = CF