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Question-1

In the quadrilateral ABCD, AC = AD and AB bisects A (see figure). Show that Δ ABC Δ ABD. What can you say about BC and BD?

Solution:
Given: In the quadrilateral ACBD, AC = AD and AB bisects A.

To prove: Δ ABC Δ ABD                               | Given
Proof: In Δ ABC and Δ ABD,      

       AB = BA                                                | Common

       AC = AD                                                | Given AB bisects A
    CAB = DAB                                          | SAS Rule
Δ ABC Δ ABD                                           | CPCT
    BC = BD

Question-2

ABCD is a quadrilateral in which AD = BC and DAB = CBA. Prove that: 
     (i) ΔABD ΔBAC
     (ii) BD = AC
     (iii) ABD = BAC

Solution:
 
Given: ABCD is a quadrilateral in which AD = BC and DAB = CBA.
To Prove: (i) ΔABD   ΔBAC
               (ii) BD = AC
               (iii) ABD = BAC
Proof: (i) In Δ ABD and Δ BAC,
      AD = BC                         [Given]
      AB = BA                         [AB is common]
  ÐDAB = CBA                    [Given]

∴ ΔABD ΔBAC                    [SAS criterion]

(ii) In ΔABD  ΔBAC
DAB = CBA
      BD = AC                      [CPCT]

(iii) ΔABD = ΔBAC
      AD = BC                         [Given]
  ABD = BAC.               [CPCT]

Question-3

AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:
Given: AD and BC are equal perpendiculars to a line segment AB.

To Prove: CD bisects AB.

Proof: In Δ OAD and Δ OBC

          AD = BC                           | Given
     OAD = OBC                     | Each = 90°
     AOD = BOC                     | Vertically Opposite Angles

ODA = OCB                      | AAS Rule
          OA = OB                           | CPCT
\CD bisects AB.

Question-4

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that Δ ABC Δ CDA.

Solution:
Given: l and m are two parallel lines intersected by another pair of parallel lines p and q.

To Prove: Δ ABC Δ CDA.

Proof: AB || DC
          AD || BC

Quadrilateral ABCD is a parallelogram. |A quadrilateral is a parallelogram if both
                                                                |the pairs of opposite sides are parallel        BC = AD              ...(1)      | Opposite sides of a parallelogram are equal

           AB = CD              ...(2)      | Opposite sides of a parallelogram are equal
And ABC = CDA         ...(3)      | Opposite angles of a parallelogram are equal

In Δ ABC and Δ CDA,
     AB = CD                      | From (2)
     BC = DA                      | From (1)
ABC = CDA                  | From (3)
Δ ABC Δ CDA.                  | SAS Rule

Question-5

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see figure). Show that:
(i) Δ APB Δ AQB
(ii) BP = BQ or B is equidistant from the arms of A.

Solution:
Given: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A.
To Prove: (i) Δ APB Δ AQB
               (ii) BP = BQ Or
B is equidistant from the arms of A.

Proof: (i) In Δ APB and Δ AQB,
BAP = BAQ                    | Since, QAP is the bisector of A
     AB = AB                         | Common
BPA = BQA                    | Each = 90°
(Since BP and BQ are perpendiculars from B to the arms of A)
Δ APB Δ AQB                  | AAS Rule

(ii) Δ APB Δ AQB                | Proved in (i) above
         BP = BQ.                    | CPCT

Question-6

In figure, AC = AE, AB = AD and   BAD =   EAC. Show that BC = DE.
 




Solution:
Given: In figure, AC = AE, AB = AD and BAD = EAC.

To Prove: BC = DE.

Proof: In Δ ABC and Δ ADE,

                      AB = AD                             | Given

                      AC = AE                             | Given

                BAD = EAC                        | Given


BAD + DAC = DAC + EAC           | Adding DAC to both sides

             BAC = DAE

              ∴ Δ ABC Δ ADE                         | SAS Rule

                 BC = DE.                            | CPCT

Question-7

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (see figure). Show that

(i) Δ DAP Δ EBP

(ii) AD = BE


Solution:
Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB.

To Prove : (i) Δ DAP Δ EBP

(ii) AD = BE.

Proof: (i) In Δ DAP and Δ EBP,

     AP = BP                           | Since P is the midpoint of the line segment AB


DAP = EBP                       | Given

EPA = DPB                       | Given


EPA + EPD = EPD + DPB         |Adding EPD to both sides


APD = BPE                      | ASA Rule

∴ Δ DAP Δ EBP                

 
(ii) Δ DAP Δ EBP                      | From (1) above

AD = BE.                         | CPCT

Question-8

In the right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that: 
(i) Δ AMC   Δ BMD
(ii)  DBC is a right angle
(iii) Δ DBC = Δ ACB
(iv) CM=  AB

 

 


Solution:
Given: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.
To Prove: (i) Δ AMC Δ BMD
               (ii) DBC is a right angle
               (iii) Δ DBC Δ ACB
               (iv) CM = AB.

Proof : (i) In Δ AMC and Δ BMD,
AM = BM                         | Since M is the mid-point of the hypotenuse AB
CM = DM                         | Given
AMC = BMD                   | Vertically Opposite Angles

Δ AMC
  Δ BMD.                  | SAS Rule
(ii) Δ AMC Δ BMD                  | From (i) above
ACM = BDM                  | CPCT

But these are alternate interior angles and they are equal
AC || BD
Now, AC || BD and a transversal BC intersects them
DBC+ ACB = 180°

| The sum of the consecutive interior angles on the same side of the transversal is 180°             DBC + 90° = 180°             | Since ACB = 90° (given)
DBC = 180° - 90° = 90°

DBC is a right angle.


(iii) In Δ DBC and Δ ACB,
          DBC = ACB (each = 90°)        | Proved in (ii) above
              BC = CB                                | Common
      Δ AMC = Δ BMD                           | Proved in (i) above
              AC = BD                                | CPCT
         Δ DBC    Δ ACB.                           | SAS Rule

(iv) Δ DBC   Δ ACB                         | Proved in (iii) above
           DC = AB                              | CPCT
      ⇒ 2CM = AB                              | DM = CM= DC
        ⇒ CM = AB.

Question-9

In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that:
(i) OB = OC (ii) AO bisects A.

Solution:

Given: In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O.

To Prove: (i) OB = OC
               (ii) AO bisects A

Proof: (i) AB = AC    | Given
          B = C   | Angles opposite to equal sides of a triangle are equal
        B = C |
Q BO and CO are the bisectors of B and C respectively
        OBC = OCB

           OB = OC         | Sides opposite to equal angles of a triangle are equal
 
(ii) In Δ OAB and Δ OAC, | Given
AB = AC                  | Given
OB = OC                 | Proved in (i) above
B = C               | Angles opposite to equal sides of a triangle are equal
B = C
∴ ∠ ABO = ACO
∴ Δ OAB  D  OAC           | Since BO and CO are the bisectors of B and C respectively
∴ ∠ OAB = OAC          | By SAS Rule
AO bisects A          | CPCT

Question-10

In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that D ABC is an isosceles triangle in which AB = AC.


 


Solution:
Given: In Δ ABC, AD is the perpendicular bisector of BC.


To Prove: Δ ABC is an isosceles triangle in which AB = AC.


Proof: In Δ ADB and Δ ADC,

    ADB = ADC                         | Each = 90°

DB = DC                               | Since AD is the perpendicular bisector of BC

AD = AD                               | Common

Δ ADB = Δ ADC                           | By SAS Rule

     AB = AC                               | CPCT

Δ ABC is an isosceles triangle in which AB = AC.

Question-11

ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:
Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively.
To Prove: BE = CF.
Proof: ABC is an isosceles triangle
     AB = AC

ABC = ACB    ...(1)| Angles opposite to equal sides of a triangle are equal
In Δ BEC and Δ CFB,
   Ð BEC = CFB             | Each = 90°
       BC = CB                  | Common
   ECB = FBC             | From (1)

Δ BEC ≅ Δ CFB              | By ASA Rule
    BE = CF.                 | CPCT

Question-12

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:
(i) Δ ABE   Δ ACF
(ii) AB = AC, i.e., Δ ABC is an isosceles triangle.

Solution:
Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.

To Prove: (i) Δ ABE Δ ACF
               (ii) AB = AC, i.e., Δ ABC is an isosceles triangle.

Proof: (i) In Δ ABE and Δ ACF
         BE = CF                   | Given
    BAE = CAF              | Common
    Ð AEB = AFC              | Each = 90°
Δ ABE   Δ ACF               | By AAS Rule

(ii) Δ ABE Δ ACF             | Proved in (i) above
     AB = AC                  | CPCT
Δ ABC is an isosceles triangle.

Question-13

 ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ABD = ACD.

Solution:
Given: ABC and DBC are two isosceles triangles on the same base BC.

To Prove: ABD = ACD                

Proof:  ABC is an isosceles triangle on the base BC
           ABC = ACB                                 …..(1) 
 DBC is an isosceles triangle on the base BC
          DBC = DCB                              …..(2)
Adding the corresponding sides of (1) and (2), we get
ABC + DBC = ACB + DCB
         ABD = ACD.

Question-14

Δ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that BCD is a right angle.


Solution:
Given: Δ ABC is an isosceles triangle in which AB = AC.

               Side BA is produced to D such that AD = AB.


To Prove: BCD is a right angle.


Proof: ABC is an isosceles triangle

ABC = ACB               ...(1)

     AB = AC and AD = AB

     AC = AD

In Δ ACD,

    CDA = ACD            | Angles opposite to equal sides of a triangle are equal

CDB = ACD               ...(2)

Adding the corresponding sides of (1) and (2), we get
 

ABC + CDB = ACB + ACD

ABC + CDB = BCD             …(3)

In Δ BCD,

    BCD + DBC + CDB =  180°        | Sum of all the angles of a triangle is 180°

BCD + ABC + CDB =  180°


            BCD + BCD = 180°          | Using (3)

                       2 BCD = 180°

                          Þ BCD = 90°

BCD is a right angle.

Question-15

ABC is a right angled triangle in which A = 90° and AB = AC. Find B and C.

Solution:

 

In Δ ABC, AB = AC

B = C      ...(1)                 | Angles opposite to equal sides of a triangle are equal

In Δ ABC,

A + B + C = 180°             | Sum of all the angles of a triangle is 180°

90° + B + C = 180°          | Since A = 90° (given)

B + C = 90°  ...(2)

From (1) and (2), we get

B = C = 45°.

Question-16

Show that the angles of an equilateral triangle are 60° each.

Solution:
Given: An equilateral triangle ABC. 

To Prove : A = B = C = 60°.

Proof: ABC is an equilateral triangle

AB = BC = CA               ...(1)
 

AB = BC

A = C                    ...(2) | Angles opposite to equal sides of a triangle are equal
 

BC = CA

A = B                   ...(3) | Angles opposite to equal sides of a triangle are equal

From (2) and (3), we obtain
 

  A = B = C           ... (4)

In Δ ABC,

A + B + C = 1800  … (5) | sum of all the angles of a triangle is 1800

From (4) and (5), we get

A = B = C = 600

Question-17

 Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that:
(i) Δ ABD Δ ACD
(ii)Δ ABP Δ ACP
(iii) AP bisects A as well as D
(iv) AP is the perpendicular bisector of BC.

Solution:
Given: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.

To Prove: (i) Δ ABD Δ ACD
               (ii) Δ ABP Δ ACP
               (iii) AP bisects A as well as D
               (iv) AP is the perpendicular bisector of BC.

Proof: (i) In Δ ABD and Δ ACD,
        AB = AC             …..(1) | Δ ABC is an isosceles triangle
        BD = CD              ...(2) | Δ DBC is an isosceles triangle
        AD = AD              ...(3) | Common
Δ ABD @ Δ ACD                  | SSS Rule

(ii) In Δ ABP and Δ ACP,
      AB = AC          ...(4)  | From (i)

  ABP = ACP ...(5) 
| AB = AC( from (1)   
 ÐABP = Ð ACP            |Angles opposite to equal sides of a triangle are equal 
Δ ABP Δ ACP             | Proved in (i) above
BAP = CAP ...(6)   | CPCT
In view of (4), (5) and (6)
   Δ ABP Δ ACP

(iii) Δ ABP Δ ACP
  ∴ ∠ BAP = CAP
AP bisects A
In Δ BDP and Δ CDP,
       BD = CD            ...(7) | From (2) 
       DP = DP            ...(8) | Common
Δ ABP = Δ ACP               | Proved in (ii) above
     \ BP = CP         ...(9)   | CPCT

In view of (7), (8) and (9),
Δ BDP = Δ CDP
∴ ∠ BDP = CDP
DP bisects D                 | SSS Rule
AP bisects D                 | CPCT
(iv) Δ BDP = Δ CDP              | Proved in (iii) above
        \ BP = CP  ...(10)        | CPCT
     BPD = CPD              | CPCT    

But BPD + CPD = 180°   | Linear Pair Axiom
   ∴ ∠ BPD = CPD = 90°   …..(11)
In view of (10) and (11),
AP is the perpendicular bisector of BC.

Question-18

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that :

(i) AD bisects BC
(ii) AD bisects A.

Solution:
Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To Prove: (i) AD bisects BC

               (ii) AD bisects A.

Proof: (i) In right Δ ADB and right Δ ADC,

Hyp. AB = Hyp. AC          | Given
Side AD = Side AD          | Common
Δ ADB   Δ ADC           | RHS Rule
    BD = CD                | CPCT
AD bisects BC.

(ii)  Δ ADB Δ ADC         | Proved in (i) above
    BAD = CAD         | CPCT
AD bisects A.

Question-19

 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of triangle PQR (see figure). Show that:
(i) Δ ABM Δ PQN
(ii) Δ ABC Δ PQR

Solution:
Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR.

To Prove: (i) Δ ABM Δ PQN
(ii) Δ ABC Δ PQR.

Proof: (i) In Δ ABM and Δ PQN,        ...(1)
AB = PQ      ...(2)   | Given
AM = PN                | Given
BC = QR                | M and N are the mid-points of BC and QR respectively

2BM = 2QN

  ⇒ BM = QN           ….(3)

In view of (1), (2) and (3),
Δ ABM Δ PQN                | SSS Rule

(ii)  Δ ABM Δ PQN          | Proved in (1) above
ABM = PQN           | CPCT
ABC = PQR              …… (4)
In Δ ABC and Δ PQR,
        AB = PQ                    | Given
        BC = QR                    | Given
   ABC = PQR                | From (4)
Δ ABC PQR                 | SAS Rule

Question-20

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given: BE and CF are two equal altitudes of a triangle ABC.

To Prove: Δ ABC is isosceles. 

Proof: In right Δ BEC and right Δ CFB, 
Side BE = Side CF           | Given
Hyp. BC Hyp. CB          | Common  
Δ BEC ≅ Δ CFB             | RHS Rule
  BCE = CBF             | CPCT
AB = AC                  | Sides opposite to equal angles of a triangle are equal
Δ ABC is an isosceles.

Question-21

ABC is an isosceles triangle with AB = AC. Draw AP BC to show that B = C.

Solution:
Given: ABC is an isosceles triangle with AB = AC.


To Prove: B = C


Construction: Draw AP BC


Proof : In right triangle APB and right triangle APC,

Hyp. AB = Hyp. AC      | Given

Side AP = Side AP       | Common

Δ APB Δ APC         | RHS Rule

   ABP = ACP           | CPCT

   B = C.

Question-22

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:
Let ABC be a right angled triangle in which B = 90°.

Then, A + C = 90° | Sum of all the angles of a triangle is 180°

  B = A + C

B > A

and B > C

AC > BC                 | Side opposite to greater angle is longer and AC > AB

AC is the longest side, i.e., hypotenuse is the longest side.

Question-23

In the figure, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.

Solution:
Given: Sides AB and AC of Δ ABC are extended to points P and Q respectively. Also,
PBC < QCB.


To Prove:   AC >AB.

Proof:   PBC < QCB             | Given
       - PBC > - QCB
180° - PBC > 180° - QCB
          ABC > ACB
               AC > AB               |Since the side opposite to the greater angle is longer

Question-24

In figure,  B <  A and  C <  D, Show that AD < BC.
 


Solution:
Given: In figure, B < A and C < D.


To Prove: AD < BC                           | Given

Proof:       B < A

            A > B

                OB > OA                  ...(1) | Side opposite to greater angle is longer

               C < D                         | Given

               D > C

               OC > OD                  ...(2) | Side opposite to greater angle is longer

From (1) and (2), we get

OB + OC > OA + OD

     BC > AD

    AD < BC.

Question-25

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A >  C and  B >  D.

 


Solution:
Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.

To Prove: A > C and B
> D

Construction: Join AC.

Proof: In Δ ABC,
                AB < BC                    | Given, AB is the smallest side of quadrilateral ABCD
             BC > AB
         BAC > BCA      ...(1) | Angle opposite to longer side is greater
In Δ ACD,    CD > AD                           | Given, CD is the longest side of quadrilateral ABCD
              ∠ CAD > ACD            | Angle opposite to longer side is greater ….(2)
From (1) and (2), we obtain
BAC + CAD > BCA + ACD



A > C
Similarly, joining B to D, we can prove that B > D.

Question-26

In figure, PR > PQ and PS bisects  QPR. Prove that  PSR >  PSQ.

 

 


Solution:
Given: In figure, PR > PQ and PS bisects QPR.

To Prove: PSR > PSQ
Proof: In Δ PQR,

        PR > PQ   (Given) 

PQR > PRQ    ...(1) (  Angle opposite to longer side is greater) 

PS is the bisector of ∠ QPR

QPS = RPS    ...(2)
In Δ PQS,
PQS + QPS + PSQ = 180° ...(3) ( The sum of the three angles of a triangle is 180°)
In Δ PRS,
PRS + SPR + PSR = 180° ...(4)

Sum of the three angles of triangle
 is 180° from (3) and (4),
PQR + QPS + PSQ = PRS + SPR + PSR
PQR + PSQ = PRS + PSR           (from (2) )
PRS + PSR = PQR + PSQ
PRS + PSR > PRQ + PSQ            (From (1))
PRQ + PSR > PRS + PSQ            (   PRQ = PRS)
PSR > PSQ

Question-27

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution:
Given: l is a line and P is a point not lying on l. PM l. N is any point on l other than M.

To Prove: PM < PN.

Proof: In Δ PMN,

M = 90°

⇒ ∠ N is an acute angle.       | Angle sum property of a triangle
⇒ ∠ M > N
PN > PM                         | Side opposite to greater angle is greater
PM < PN.

Question-28

ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of Δ ABC.

Solution:
Construct the perpendicular bisectors of two sides of Δ ABC. Their point of intersection is the required point.

Question-29

In a triangle locate a point in its interior which is equidistant from all the sides of triangle.

Solution:
The angle bisector between of any two angles of the triangle. Their point of intersection is the required point which is the incentre. Incentre is equidistant from all the sides of triangle.

Question-30

[Hint: The parlour should be equidistant from A, B and C]

 



Solution:
Draw the perpendicular bisectors of AB and AC. Their point of intersection is the required point.

Question-31

Complete the hexagonal and star shaped Rangolies [see figures (i) and (ii)] by filling them with as many equilateral triangles of side l cm as you can. Count the number of triangles in each case. Which has more triangles?

             (i)                                (ii)

Solution:
(i) Number of triangles = 25 + 25 + 25 + 25 + 25 + 25 = 150

(i)


(ii) Number of triangles = 24 × 12 = 288.
 

(ii)

Figure (ii) has more triangles.





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