Triangle Class 9

 Triangle

Exercise 

Q 1 ) In ∆ ABC, AB = AC, D is a point inside the triangle such that ∠DBC = ∠ DCB. Prove that ∆ BAD ≅ ∆CAD.

Solution :

Given : In ∆ ABC, AB = AC

            D is a point inside the triangle such that ∠DBC = ∠ DCB.

To prove : ∆ BAD ≅ ∆CAD.

Proof : In ∆ DBC

            ∠DBC = ∠ DCB ( Given)

            DC = DB ( sides opposite to equal angles)

            Now in ∆ BAD and ∆CAD

             AB = AC           ( Given)

             AD = AD           ( Common)     

             DB = DC           ( Base angles are equal in ∆BDC ( ∠DBC = ∠ DCB))

         ∴  ∆ BAD ≅ ∆CAD ( SSS rule of congruency)

Q 2) In ∆ PQR, PQ = PR, QN = RM. Prove that ∠ QPM = ∠RPN.

Solution : 

Given : In ∆ PQR, PQ = PR, QN = RM.

To Prove : ∠ QPM = ∠RPN.

Proof : In ∆ PQR

QN = RM ( Given)

(Substracting MN from both sides)

QN - MN = RM - MN

 (QM = NR)

Now in ∆ PQM and ∆ PRN; we have

PQ = PR     ( Given)

∠Q = ∠R     ( Opposite to equal sides)

QM = NR   ( Proves above)

 ∴  ∆ PQM ≅ ∆PRN ( SAS rule of congruency)

 

∴   ∠ QPM = ∠RPN   ( c.p.c.t)

Q 3) In ∆ ABC, AB = AC , and CP = BQ are altitudes. Prove that CP = BQ

Solution:

Given : In ∆  ABC,

AB = AC,  CP and BQ are altitude on AB and AC respectively.

To given : CP - BQ

Proof : 

In ∆ ABC,

AB = AC      ( Given )

∠B = ∠C      ( Opposite to equal side )

Now in ∆PBC and ∆ QBC

∠P = ∠ Q             ( Each 90°)

∠B = ∠C              ( Prove above)

BC = BC              ( common )

 ∴  ∆ PBC ≅ ∆QBC ( AAS rule of congruency)

 ∴  CP = BQ     ( c.p.c.t)

Q 4) In the quadrilateral ABCD, AD = CD and ∠A = 90° = ∠C. Prove that AB = BC .

Solution : 

Given: In quadrilateral ABCD 

AD = CD 

∠A = 90° = ∠C

To prove : AB = BC

Proof : In right  ∆ ABD and  ∆ CBD

 ∠BAD  = ∠BCD        ( each  90°)

    BD    =  BD             ( common)

side AD = CD             ( Given)

∴  ∆ ABD ≅ ∆CBD ( RHS rule of congruency)

∴  AB = BC      ( c.p.c.t)

Q 5) In the given triangle, AC = DF, BD = CE and ∠ACB = ∠ FDE. Prove that ∠A = ∠F.

Solution: 

Given :

In ∆ABC and ∆DEF, AC = DF, BD = CE

∠ACB = ∠FDE

 

To prove  : ∠A = ∠ E

Proof : 

BD = CE            ( Given)

Adding CD to both sides

BD + CD =  CD + CE

BC = DE 

Now in ∆ABC and ∆DEF

AC = DF             ( Given )

BC = DE             ( Proved)

∠ACB = ∠FDE    ( Given)

∴  ∆ ABC ≅ ∆DEF ( SAS rule of congruency)

∠A = ∠ E                 ( c.p.c.t)

Q 6) In the given figure, ∆ABC is right angled at B, ACDE and BCGF are squares. Prove that

i) ∆ BCD ≅ ∆ ACG

ii) AG = BD.

Solution :

Given : In the given figure, ∆ ABC is a right angled at B. 

ACDE and BCGF squares on AC and BC respectively.  BD and AG are joined.

To prove : i) ∆ BCD ≅ ∆ ACG

                 ii) AG = BD.

Proof : 

∠ BCA + ∠ACD = ∠BCA + ∠ BCG    { ∠ACD  and ∠ BCG are each  90°}

∠BCD = ∠ACG

Now in ∆ BCD and ∆ ACG; we have

BC = CD                (sides of square)

∠BCD = ∠ACG     ( Proved)

CD = AC                ( Sides of a square)

∴  ∆ BCD ≅ ∆ACG ( SAS rule of congruency)

∴  BD = AG ( c.p.c.t)

or AG = BD

Q 7) Diagonal AC is the perpendicular bisector of diagonal BD in the quadrilateral ABCD . Prove that 

                 i) AB = AD 

                 ii) BC = DC 

Solution : 

To prove : i) AB = AD 

                 ii) BC = DC 

Proof : AC is perpendicular bisector of BD at O

           ∴ BO = OD

and ∠AOB = ∠AOD = 90°

Now in ∆AOB and ∆ AOD

BO = OD               ( Proved)

AO =AO                ( Common)

∠AOB = ∠AOD    ( each 90°)

∴  ∆ AOB ≅ ∆AOD ( SAS rule of congruency)

    AB = AD ( c.p.c.t)

Similarly in  ∆ BOC and   ∆ DOC

       OC = OC            ( Common)

       BO = OD            ( Proved)

   ∠BOC = ∠DOC     ( each 90°)

∴  ∆ BOC ≅ ∆DOC  ( SAS rule of congruency)

∴ BC = DC               ( c.p.c.t)

Q 8) PS bisect ∠QPR and PS ⊥ QR. If PQ = 2x units, PR = (3y + 8) units, QS = x units and SR = 2y units. Find the values of x and y.

Solution:

Given .

PS bisects ∠QPR and PS ⊥ QR

PQ = 2x PQ = 2x units, PR = (3y + 8) units, QS = x units and SR = 2y units.

To find : The values of x and y.

In ∆PQS and ∆PRS

PS = PS                 ( Common)

 ∠QPS =  ∠RPS ( PS is the bisector of  ∠P)

 ∠PSQ =  ∠PSR     (  PS ⊥ QR)

∴  ∆ PQS ≅ ∆PRS  ( SAS rule of congruency)

PQ = PR and QS = RS ( c.p.c.t)

Now 2x = 3y + 8 

 2x - 3y = 8 -----------(1)

and x = 2y  ------------(2)

Putting 2 in 1

2(2y) - 3y = 8

4y - 3y = 8

         y = 8 units

x = 2y

x = 2 * 8

x = 16 units

Q 9. In ∆AOB and ∆COD,  ∠ B = ∠ C and O is the mid point of BC. 

Find the values of x and y if AB = 3x units. CD = y + 2 units, AO = x + 2 units

DO = y units.

Solution :

Given : In ∆AOB and ∆COD,  

∠ B = ∠ C and O is the mid point of BC i.e. BO = OC

AB = 3x units. CD = y + 2units, 

AO = x + 2 units

DO = y units.

To find : The value of x and y.

In ∆AOB and ∆COD,  

BO = OC            ( Given)

∠  AOB = ∠ COD   ( Vertically opposite angles)

∠ B = ∠ C                (Given)

∴  ∆ AOB ≅ ∆COD  ( ASA rule of congruency)

AB = CD and AO = CO   ( c.p.c.t)

∴  3x = y + 2

      y = 3x -2     ----------------------(i)

   x + 2 = y ----------------------------(ii)

From ( i) and (ii)

3x - 2 = x + 2

3x - x = 2 + 2

2x = 4

x = 4/2 = 2

y = x + 2 = 2 + 2 = 4

Hence, x =2 ; y = 4

Q 10) Prove that ∆ABD ≅ ∆CBD. Find the values of x and y, if ∠ABD = 35°, ∠CBD = (3x + 5)°,  

∠ADB = (y - 3)°. ∠CDB = 25°.

Solution .

Given :

In ∆ABD and ∆CBD

AB = BC and AB = CD

∠ABD = 35°, ∠CBD = (3x + 5)°,  

∠ADB = (y - 3)°. ∠CDB = 25°.

To prove : 

In ∆ABD ≅ ∆CBD, we have

AB = BC                       ( Given)

AD = CD                       ( Given)

BD = BD                       ( Common)

∴  ∆ ABD ≅ ∆CBD         (SAS rule of congruency)

∴ ∠ABD = ∠CAD and ∠ADB = ∠CDB           (c.p.c.t)

∴  35° = (3x + 5)°

∴  35° -  5° = 3x 

              x = 30/ 3

              x = 10°

and y - 3° = 25°

       y = 25° + 3°

       y = 28°