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Chapter 7. Triangles

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Exercise 7.1 

Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD.

Solution:

Given: AC = AD and AB bisects ∠A

To prove:  Δ ABC ≅ Δ ABD.

Proof: In Δ ABC and Δ ABD.                   

                      AC = AD       [given]               

                  ∠CAB = ∠BAD   [AB bisect ∠A]      

                       AB = AB         [Common]

By SAS Congruence Criterion Rule

                 Δ ABC ≅ Δ ABD

                      BC = BD [By CPCT]   Proved

Q2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig.7.17). Prove that

(i) Δ ABD ≅ Δ BAC

(ii)   BD = AC

(iii)  ∠ ABD = ∠ BAC                                                                                        

Solution:

Given: ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA

To prove: 

(i) Δ ABD ≅ Δ BAC

(ii)   BD = AC                    

(iii)  ∠ ABD = ∠ BAC

Proof:  (i) In Δ ABD and Δ BAC

                         AD = BC        [given]

                   ∠ DAB = ∠ CBA   [given]

                         AB = AB         [Common]

  By SAS Congruency Criterion Rule

                  Δ ABD ≅ Δ BAC

(ii)                    BD = AC [CPCT]

(iii)            ∠ ABD = ∠ BAC    [CPCT]

Q3.  AD and BC are equal perpendiculars to a line segment AB (see the given 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 ∆BOC and ∆AOD

∠ BOC = ∠AOD (Vertically opposite angles)

 ∠CBO = ∠DAO (Each 90º)   

      BC = AD (Given)

By AAS Congruence Criterion Rule

  ∆BOC ≅ ∆AOD

       BO = AO (By CPCT)

Hence, CD bisects AB.

Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q (See the given figure). Show that ∆ABC ≅ ∆CDA

Solution:

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

To prove: ∆ABC ≅ ∆CDA

Proof:   

In ∆ABC and ∆CDA,       

∠ BAC = ∠DCA (Alternate interior angles, as p || q)

     AC = CA (Common)

∠ BCA = ∠DAC (Alternate interior angles, as l || m)

By AAS Congruence Criterion Rule  

∆ABC ≅ ∆CDA

Q5. 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 the given figure). Show that: (i) ∆APB ≅ ∆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) ∆APB ≅ ∆AQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Proof:  

In ∆APB and ∆AQB,

∠ APB = ∠AQB (Each 90º)

∠ PAB = ∠QAB (l is the angle bisector of A)

      AB = AB (Common)

By AAS Congruence Criterion Rule

 ∆APB ≅ ∆AQB

     BP = BQ    [CPCT]

it can be said that B is equidistant from the A.

Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

Given: AC = AE, AB = AD and ∠BAD = ∠EAC.

To prove: BC = DE.

Proof:  ∠BAD = ∠EAC   

   BAD + DAC = EAC + DAC

               BAC = DAE

    In ∆BAC and ∆DAE

             AC = AE (Given)

             AB = AD (Given)

           ∠BAC = ∠DAE (proved above)

By SAS Congruence Criterion Rule

          ∆BAC ≅ ∆DAE

               BC = DE (CPCT)

Q7.  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 the given figure).

Show that: 

(i) ∆DAP ≅ ∆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) ∆DAP ≅ ∆EBP

(ii) AD = BE

Proof:  In ∆ DPA and ∆ EPB

∠EPA = ∠DPB

      EPA + DPE = DPB + DPE

             ∠ DPA = ∠EPB

      

        ∠BAD =∠ ABE (Given)

 

         ∠EPA = ∠DPB (Given)

 

             AP =BP (P is the midpoint of AB)

 

    By AAS Congruence Criterion Rule

 

          ∆DAP ≅ ∆EBP

                  AD = BE (CPCT)

 

 

 

 

Chapter 7. Triangles

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Exercise 7.2

Q1. 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.

To prove:

(i) OB = OC

(ii) AO bisects ∠ A

Proof: In  ΔABC, We have:

           AB = AC 

          ∠ B = ∠ C [ opposite angle to the equal side ]

Or  1/2 ∠ B = 1/2 ∠C

So, ∠OBC = ∠OCB […1]

In D ABO and D ACO

     AB = AC [given]

∠OBC = ∠OCB [from …1]

     AO = AO [common]

By SAS Congruence Criterion Rule

DABO ≅  DACO

OB = OC [ By CPCT ]

∠BAO = ∠CAO [ By CPCT ]

AO bisect ∠A.

Q2. In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ 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 ABD and Δ ACD, we have

       DB = DC [since D bisect BC]

 ∠ BDC = ∠ADC [AD is the perpendicular bisector of BC].

       AD = AD [common]

By SAS Congruence Criterion Rule

    Δ ABD ≅ Δ ACD

          AB =AC [CPCT]

Hence, Δ ABC is an isosceles triangle

Q3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Solution: 

Given: ABC is an isosceles triangle in which BE ⊥ AC and CF ⊥ AB where AB = AC.

To prove: BE = CF.

Proof:  Here, BE ⊥ AC and CF ⊥ AB  (Given) 

         In ΔABE and Δ ACF

            ∠ AEB = ∠ AFC  (90^० Each) 

                 ∠ A = ∠ A       (Common)

                  AB = AC        (Given) 

By ASA Congruency Criterion Rule

         ΔABE  ≅  Δ ACF

               BE = CF [ By CPCT ]

Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Solution: 

Given: ABC is a triangle in which 

BE ⊥ AC and CF ⊥ AB and BE = CF 

To Prove : 

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Proof : 

(i) In Δ ABE and Δ ACF

                BE = CF           (Given) 

          ∠ AEB = ∠ AFC     (90^० Each) 

               ∠ A = ∠ A         (Common)

Using ASA Congruency Property 

          Δ ABE ≅ Δ ACF   Proved I 

(ii)            AB = AC  [By CPCT] 

Therefore, ABC is an isosceles triangle. 

Q5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). 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 in which 

               AB = AC      (Given) 

 ∴      ∠ ABC = ∠ ACB           .......... (1) 

      (Opposite angles of equal sides) 

Similarily, 

       BCD is also an isosceles triangle. 

               BD = CD      (Given) 

 ∴      ∠ DBC = ∠ DCB           .......... (2) 

      (Opposite angles of equal sides) 

Adding Equation (1) and (2) 

 ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB

Or,           ∠ ABD = ∠ ACD       Proved

Q6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.

Solution : 

Given: ΔABC is an isosceles triangle in which AB = AC.

Side BA is produced to D such that AD = AB 

To prove : ∠ BCD = 90^०

Proof : 

           AB = AC      .............. (1)  (Given) 

And,   AB = AD      .............. (2)  (Given) 

From equation (1) and (2) we have 

         AC = AD       ...............(3) 

 ∴      ∠3 = ∠4     .... (4) (Opposite angles of equal sides) 

Now, AB = AC    from (1) 

∴       ∠1 = ∠2     .... (5) (Opposite angles of equal sides)

In ΔABC, 

Exterior ∠5 = ∠1 + ∠2    

Or,         ∠5 = ∠2 + ∠2       from (5) 

Or,         ∠5 = 2∠2    ....... (6)

Similarily, 

Exterior ∠6 = ∠3 + ∠4 

Or,         ∠6 = 2∠3      from (7) 

Adding equation (6) and (7) 

    ∠5 + ∠6  = 2∠2 + 2∠3 

    ∠5 + ∠6  = 2(∠2 + ∠3)  

Or,     180^० = 2(∠2 + ∠3)   [ ∵ ∠BAC + ∠DAC  = 180^० ]

Or,    ∠2 + ∠3 = 180^० / 2

Or,     ∠BCD = 90^०      Proved

Q7. ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Solution: 

Given : ABC is a right angled triangle in which

∠ A = 90° and AB = AC.

To Find : ∠B and ∠C  

              AB = AC          (Given) 

∴            ∠B = ∠C      ............(1)

(Opposite angles of equal sides)

In triangle ABC, 

∠A + ∠B + ∠C = 180^०   (Angle sum property) 

90° + ∠B + ∠B = 180^०   Using equation (1) 

2 ∠B = 180^० - 90° 

2 ∠B = 90°

   ∠B =  90°/ 2 

   ∠B =  45°

∴ ∠B = 45° and ∠C =  45° 

Q8. Show that the angles of an equilateral triangle are 60° each. ⊥ 

Solution: 

Given: ABC is a equilateral triangle, in which 

AB = BC = AC 

To Prove : 

∠A = ∠B = ∠C = 60°

Proof :

AB = AC      (Given) 

∠B = ∠C     ....................... (1)   [opposite angle of equal sides]

AB = BC       (Given) 

∠A = ∠C     ....................... (2)   [opposite angle of equal sides]

AC = BC       (Given) 

 

∠A = ∠B     ....................... (3)   [opposite angle of equal sides]

From equation (1), (2) and (3) we have 

 ∠A = ∠B = ∠C     .............. (4) 

In triangle ABC 

∠A + ∠B + ∠C = 180°

∠A + ∠A + ∠A = 180°

3 ∠A = 180°

   ∠A = 180°/3 

   ∠A = 60° 

∴ ∠A = ∠B = ∠C = 60° 
 

EXERCISE- 7.3

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 1.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 Fig. 7.39). 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.

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:

In ΔABD and Δ ACD

              AB = AC [given]

           BD = CD [given]

           AD = AD [common]

   By SSS Congruence Criterion Rule

         Δ ABD ≅ Δ ACD

     ∠ BAD = ∠CAD [CPCT]

       ∠ BAP = ∠CAP [CPCT] … 

 (ii)In ΔABP and Δ ACP

           AB = AC [given]

       ∠ BAP = ∠CAP [proved above]

             AP = AP [common]

   By SAS Congruence Criterion Rule

             Δ ABP ≅ Δ ACP

                   BP = CP [CPCT] … 2

                    ∠APB = ∠APC [CPCT]

(iii)      ∠ BAP = ∠CAP [From eq. 1]

                Hence, AP bisects ∠ A.

 Now, In Δ BDP and Δ CDP

BD = CD [given]

      BP = CP [given]

      DP = DP [common]

By SSS Congruence Criterion Rule

      Δ BDP ≅ Δ CDP

    ∠ BDP = ∠CDP [CPCT]

AP bisects ∠ D.

(iv) AP stands on B

∠APB + ∠APC = 180⁰

∠APB +∠APB = 180⁰[proved above]

    ∠APB = 180⁰  /2

 ∠APB = 90⁰

AP is the perpendicular bisector of BC.

2. 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: In ∆BAD and ∆CAD 

  ∠ ADB = ∠ADC (Each 90º as AD is an altitude)

     AB = AC (Given)

     AD = AD (Common)

By RHS Congruence Criterion Rule

∆BAD ≅ ∆CAD 

BD = CD (By CPCT)

Hence, AD bisects BC. 

∠BAD = ∠CAD (By CPCT)

Hence, AD bisects ∠ A

3. Two sides AB and BC and median AM of one triangle ABC are respectively

equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). 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 Δ PQR.

To prove: (i) Δ ABM ≅ Δ PQN

               (ii) Δ ABC ≅ Δ PQR

Proof: In ∆ABC, AM is the median to BC.

BM = 1/2 BC ... 1

In ∆PQR, PN is the median to QR.

QN = 1/2 QR ... 2

from eq .1 & 2 

BM = QN ... 3

Now in ABM and  PQN

AB = PQ (Given)

BM = QN [From equation (3)]

AM = PN [given]

By SSS congruence Criterion rule

∆ABM ≅ ∆PQN 

∠B =∠Q [CPCT]

Now in∆ ABC and∆ PQR 

AB = PQ [given]

∠B = ∠Q [prove above ]

BC = QR [given]

By SAS congruence Criterion rule

∆ ABC ≅ ∆ PQR 

4. 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: ABC is a isosceles.

Proof: In ∆BEC and ∆CFB,

                BE = CF (Given)

            ∠BEC = CFB (Each 90°)

                BC = CB (Common)

    By RHS congruence Criterion rule

         ∆BEC ≅ ∆CFB

    ∠BCE = ∠CBF (By CPCT)

AB = AC [Sides opposite to equal angles of a triangle are equal]

 Hence, ABC is isosceles.

5. 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 to

Proof :  In ∆APB and ∆APC

∠APB = ∠APC (Each 90º)

AB =AC (Given)

AP = AP (Common)

By RHS Congruence Criterion Rule

∆APB ≅  ∆APC

∠B = ∠C [CPCT]

Q1: Show that in a right angled triangle, the                           hypotenuse is the longest side.

Solution:

Given:Let us consider a right-angled triangle ABC, right-angled at B.

To prove: AC is the longest side.

 Proof: In ΔABC,

∠ A + ∠ B + ∠ C = 180° (Angle sum property of a 

∠ A + 90º + ∠ C = 180°

∠ A + ∠ C = 90°

Hence, the other two angles have to be acute (i.e., less than 90º).

∴  ∠ B is the largest angle in ΔABC.

⟹∠ B > ∠ A and ∠ B > ∠C

⟹ AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]

Therefore, AC is the largest side in ΔABC.

However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

 Q2 :  In the given figure sides AB and AC of ΔABC are extended to points          P and Q respectively. Also, ∠ PBC < ∠ QCB. Show that AC > AB.

Answer : Given: AB and AC of ΔABC are extended to points PandQ respectively. Also, ∠PBC < ∠QCB.

To prove: AC > AB

Proof: In the given figure,

∠ ABC + ∠ PBC = 180° (Linear pair)

⇒ ∠ ABC = 180° - ∠ PBC ... (1)

Also,

∠ ACB + ∠ QCB = 180°

∠ ACB = 180° - ∠ QCB … (2)

As ∠ PBC < ∠ QCB,

⇒ 180º - ∠ PBC > 180º - ∠ QCB

⇒ ∠ ABC > ∠ ACB [From equations (1) and (2)]

⇒ AC > AB (Side opposite to the larger angle is larger.)

Q3 :  In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Solution: 

Given:  ∠ B < ∠ A and ∠ C < ∠ D.

To prove: AD < BC

Proof: In ΔAOB,

∠ B < ∠ A

⇒ AO < BO (Side opposite to smaller angle                       is smaller... (1)

In ΔCOD,

∠ C < ∠ D

⇒ OD < OC (Side opposite to smaller angle is smaller) ... (2)

On adding equations (1) and (2), we obtain

AO + OD < BO + OC

AD < BC

Q4 :AB and CD are respectively the smallest and longest sides of a         quadrilateral ABCD (see the given 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: Let us join AC.

Proof: In ΔABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠ 2 < ∠ 1 (Angle opposite to the smaller side is smaller) ... (1)

In ΔADC,

AD < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠ 4 < ∠ 3 (Angle opposite to the smaller side is smaller) ... (2)

On adding equations (1) and (2), we obtain

∠ 2 + ∠ 4 < ∠ 1 + ∠ 3

⇒ ∠ C < ∠ A

⇒ ∠ A > ∠ C

Let us join BD.

In ΔABD,

AB < AD (AB is the smallest side of quadrilateral ABCD)

∴ ∠ 8 < ∠ 5 (Angle opposite to the smaller side is smaller) ... (3)

In ΔBDC,

BC < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠ 7 < ∠ 6 (Angle opposite to the smaller side is smaller) ... (4)

On adding equations (3) and (4), we obtain

∠ 8 + ∠ 7 < ∠ 5 + ∠ 6

⇒ ∠ D < ∠ B

⇒ ∠ B > ∠ D

Q5 : In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR               >∠ SQP.

Solution : 

Given:  PR > PQ and PS bisects ∠ QPR.

To prove: ∠ PSR >∠ SQP.

 Proof: As PR > PQ,

∴ ∠ PQR > ∠ PRQ (Angle opposite to larger side is larger) ... (1)

PS is the bisector of ∠ QPR.

∴∠ QPS = ∠ RPS ... (2)

∠ PSR is the exterior angle of ΔPQS.

∴ ∠ PSR = ∠ PQR + ∠ QPS ... (3)

∠ PSQ is the exterior angle of ΔPRS.

∴ ∠ PSQ = ∠ PRQ + ∠ RPS ... (4)

Adding equations (1) and (2), we obtain

∠ PQR + ∠ QPS > ∠ PRQ + ∠ RPS

⇒ ∠ PSR > ∠ PSQ [Using the values of equations (3) and (4)]

Q6 :  Show that of all line segments drawn from a given point not on it, the          perpendicular line segment is the shortest.

Solution: 

Given: PNM is a right angled triangle at N.

To prove: PN < PM.

Proof: In ΔPNM,

∠ N = 90º

∠ P + ∠ N + ∠ M = 180º (Angle sum property of a triangle)

∠ P + ∠ M = 90º

Clearly, ∠ M is an acute angle.

∴ ∠ M < ∠ N

⇒ PN < PM (Side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.

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