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Question-1

The bisector of the exterior angle A of Δ ABC intersects side BC produced at D. Prove that .
                      
Solution:
Given: ABC is a triangle; AD is the exterior bisector of A and meets BC produced at D; BA is produced to F.

To prove: 
               

Construction: Draw CE||DA to meet AB at E.

Proof: In Δ ABC, CE || AD cut by AC.
   CAD = ACE (Alternate angles)

Similarly CE || AD cut by AB
   FAD = AEC (corresponding angles)

Since FAD = CAD (given)
     
ACE = AEC

AC = AE ( by isosceles Δ theorem)

Now in Δ BAD, CE || DA
AE = DC (BPT)
AB    BD

But AC = AE (proved above) 

AC = DC
   AB     BD    
or

 (proved).

Question-2

State and prove the converse of angle bisector theorem.

                                

Solution:
Given: ABC is a Δ ; AD divides BC in the ratio of the sides containing the angles A to meet BC at D.
  i.e.
 
To prove: AD bisects A.
Construction: Draw CE || DA to meet BA produced at E.
Proof: In Δ ABC, CE || DA cut by AE.
    
BAD = AEC (corresponding angle) ----(i)
Similarly CE || DA cut by AC
\  DAC = ACE (alternate angles) ----(ii)
In DBEC; CE || AD
\ AB = BD (BPT)
   AE    DC
But AB = BD (given)
     AC    DC
\  AB = AB
   AE     AC
\ AE = AC
 AEC = ACE (isosceles property) ----(iii)
According to equation (i), (ii) and (iii) BAD = DAC
⇒ AD bisects A.

Question-3

If a parallelogram has all its sides equal and one of its diagonal is equal to a side, show that its diagonals are in the ratio : 1.

Solution:

Given: ABCD is a parallelogram, where AC and BD are the diagonals meeting at O. AB = BC = AC.
 
To Prove: BD : AC ::  : 1
Proof: In Δ ABC, AB = BC = CA (given).
                                    = a (say)
Hence ABC is an equilateral triangle. (Definition of equilateral triangle)
AC and BD are the diagonals of parallelogram ABCD,
AC = BD (Diagonals of a parallelogram bisect each other)
or AO = OC.
i.e BO is the median of the equilateral ABC.
Hence BO =a
BD = a   
BD : AC :: a : a
BD : AC ::  : 1.

Question-4

In Δ ABC, P, Q are points on AB and AC respectively and PQ || BC. Prove that the median AD bisects PQ.
                              

Solution:
Given: ABC is a triangle, PQ || BC; AD is the median which cuts PQ at R.
To prove: AD bisects PQ at R.
Proof: In Δ ABD; PR || BD
   AP = AR  (BPT)
   PB    RD

In Δ ACD, RQ || DC
AR = AQ (BPT)
   RD    QC

In ΔAPR and ΔABD,
APR = ABD (corresponding angles.)
ARP = ADB (corresponding angles.)
Δ APR is similar to Δ ABD (AA similarity)

AP = AR = PR (corresponding sides of similar
 triangles are proportional)----(i) 
   AB    AD    BD

Similarly Δ ARQ is similar to Δ ADC
\  AQ = AR = RQ -----(ii)
    AC   AD    DC

According to equation (i) and (ii),
   AR = PR = RQ
   AD   BD    DC
 but BD = DC (given)
PR = RQ
or AD bisects PQ at R (proved).

Question-5

Prove that the line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.

                          

Solution:
Given: In ΔABC, D, E, F are the midpoints of AB, BC and AC respectively.
 
To prove: ΔABC ~ ΔDEF
ΔABC ~ ΔADF
ΔABC ~ ΔBDE
ΔABC ~ ΔEFC

Proof: In ΔABC, D and F are mid points of AB and AC respectively. 
DF || BC (midpoint theorem)
In Δ ABC and Δ ADF
A is common; ADF = ABC (corresponding angles)
ΔABC ~ Δ DF (AA similarity) -----(1)

Similarly we can prove ΔABC ~ ΔBDE (AA similarity)-----(2)
ΔABC ~ ΔEFC (AA similarity)-----(3)
In ΔABC and ΔDEF;
since D,E, F are the midpoints of AB, BC and AC respectively,
DF= (1/2) × BC; DE = (1/2) × AC; EF = (1/2) × AB; (midpoint theorem )
AB = BC = CA = 2
   EF    DF    DE
Δ ABC ~ Δ EFD (SSS similarity )------------------(4)
From (1), (2), (3) and (4)

Δ ABC ~ Δ DEF
Δ ABC ~ Δ ADF
Δ ABC ~ Δ BDE
Δ ABC ~ Δ EFC.

Question-6

In triangle ABC, DE||BC and AD : DB = 2 : 3. Determine the ratio of the area triangle ADE to the area triangle ABC.

Solution:
==

Question-7

One angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite sides in the same ratio. Prove that the triangles are similar.
                                

Solution:
Given: ΔABC and ΔPQR
      A = P
AD and PS bisects A and P respectively.
      BD = QS
      DC    SR
To prove: ΔABC ~ ΔPQR

Proof: In ΔABC and ΔPQR
AD bisects A
AB =  BD (Angle bisector theorem) -----(1)
   AC     DC
Similarly in ΔPQR,
PQQS (Angle bisector theorem) ----(2)
PR     SR
But BD = QS (given) 
     DC    SR
According to equation (1) and (2) 
     AB = PQ 
Þ AB = AC
     AC    PR     PQ    PR
A = P (given)

ΔABC ~ ΔPQR (SAS similarity).

Question-8

The bisector of interior angle A of a triangle AB meets BC in D and the bisector of exterior angle A meets BC produced in E. Prove that
                                          

Solution:
Given: ΔABC, AD bisects interior A and AE bisects exterior A meeting BC at D and BC produced at E.
 To prove:
 
 
Proof: In ΔABC, AD bisects interior
A
∴ (Angle Bisector theorem)…………………(1)
Similarly in DABC, AE bisects exterior A
………………..(2)
From equation (1) and (2),
  
 
Hence Proved.

Question-9

In a triangle ABC, XY || AC divides the triangle into two parts equal in areas. Determine .

Solution:
Given: ABC is a triangle with XY || AC divides the triangle into two parts equal in areas.
To find:
Proof:
ar
ΔBXY = ar trap. XYCA (Given) ar ΔBXY = ar ΔABC
In
ΔBXY andBAC,
BXY = BAC (Corresponding angles)
BYX = BCA (Corresponding angles)
ΔBXY ΔBAC (AA similarity)
= (Areas of similar triangle)
=
=
AB – BX = 2 BX – BX
AX = ( 2 – 1)BX
= =.

Question-10

If two sides and a median bisecting the third side of a Δ are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar. 

                        

Solution:
Given: ΔABC and ΔPQR where BD and QS are the medians and AB = BC = BD
                                                                                    PQ    QR    QS
To prove: ΔABC ~ ΔPQR
Construction: Produce BD and QS to E and T respectively such that BD = DE and QS = ST. CE and TR are joined.

Proof: In ΔADB and
DCDE,
AD = DC (given)
ADB = CDE (Vertically opposite angles)
BD = DE.
ΔADB ΔCDE (SAS axiom)
Hence AB = CE and
ABD = DEC.

Similarly ΔPQS Δ RST,
hence PQ = TR and
PQS = STR.
Consider Δ EBC and Δ TQR,
 BD = 2 BD = BE (from given and construction)-------(1)
 QS    2QS    QT
AB = CE and PQ = RT (proved),
       AB = CE ----(2)
       PQ    RT
       AB = BC = BD  (Given)                         -----(3)
       PQ    QR    QS

From (1),(2) and (3),
      BE = CE  = BC
      QT    RT    QR
ΔEBC ~ ΔTQR(SSS similarity axiom).
DBC = SQR and DEC = STR ---(4) (corresponding angles of similar triangles are proportional)
But
ABD = DEC and PQS = STR (proved)-----(5)
ABD = PQS (from (4) and (5)) ----(6)
From (5) and (6),

ABC = PQR -----I
In Δ ABC and Δ PQR,
     AB = BC (given)
     PQ    QR
And
ABC = PQR (from I)
ΔABC ~ ΔPQR (SAS Similarity).

Question-11

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same sides of BC. If AC and DB intersect at P, prove that AP × PC = BP × PD.
                                       

Solution:
Given: Two right triangles ABC and BDC on the same hypotenuse BC. AC and BD intersect at P.

To prove: AP × PC = BP × PD

Proof: In ΔABP and ΔDCP
 
A = D (= 90°) (given)
 
APB = DPC (vertically opposite angles)
  ΔABP ~ ΔDCP (AA similarity axiom)
  AB = BP = AP  (corresponding sides of similar Δ s are
 proportional)………………….(1) 
    DC   CP     DP                     
From (1) BP = AP
            CP    DP
By cross multiplication,
BP × DP = AP × PC (proved).

Question-12

If ABC is an equilateral triangle of side 2a prove that the altitude AD = 3a and 3AB2 = 4AD2.

Solution:
 

Given: Δ ABC is an equilateral triangle of side 2a. AD is the altitude of the triangle.
To Prove: AD = a 3 and 3AB2 = 4AD2
Proof:
In rt.
Δ ADC,
AD2 = AC2 - DC2
      = (2a)2 - a2
      = 4a2 - a2
      = 3a2

AD = a3
3AB2 = 3(2a)2
       = 3(2a)2
       = 3
´ 4a2
       = 4(a
3)2
       = 4AD2

3AB2 = 4AD2.

Question-13

Two isosceles Δs have equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding heights (altitudes).

                          

Solution:
Given: ΔABC and ΔPQR are isosceles and A = P. AD, PS are the altitudes and .
To find: AD
            PS
Proof: In Δ ABC,
B = C (isosceles Δ property)

Similarly in Δ PQR, Q = R.
                  
A = P (given)

B = C =
Since
A = P
B = C = Q = R
Δ ABC ~ Δ PQR (AA)
If 2 triangles are similar then the ratio of areas will be equal to the square of the corresponding sides, 

…………………(1)
In Δ ABD, Δ PQS

D = S (= 90° )
B = Q (given)
Δ ABD ~ Δ PQS (AA)
  AB = AD 
 ……………….(2) 
    PQ    PS
According equation (1)
 
AD  =  3
    PS     4
... The ratio of their corresponding heights is 3 : 4.


Question-14

The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.

Solution:

Given: ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at ‘O’.
To prove: =
Proof:      
In the figure consider the triangle OAB and OCD
  
DOC = ∠ΑOB   (Vertically opposite angles are equal)
since AB || DC,
DCO = OAB   (Alternate angles are equal) 
By AA corollary of similar triangles.
Δ OAB Δ OCB When the two triangle are similar, the side are proportionally.
=
Hence proved.

Question-15

P and Q are the points on the sides AB and AC respectively of a ΔABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cms, QC = 6 cm, prove that BC = 3PQ.
                               

Solution:
Given: Δ ABC, PQ are points on AB and AC such that AP = 2 cm, BP = 4 cm, AQ = 3 cm, QC = 6 cm
To prove:  BC = 3PQ
  
Proof: In Δ ABC,
                       
As   AP = AQ
      PB     QC
According to converse of BPT, PQ || BC
    
In Δ APQ and Δ ABC
APQ = ABC (Corresponding angles)
A is Common
Δ APQ ~ Δ ABC (AAS similarity)
(corresponding sides of similar Δ s are proportional)

But
PQ = 2 = 1
   BC    6    3
3PQ = BC (Proved).

Question-16

Through the midpoint of M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.

                                 

Solution:
Given: ABCD is a parallelogram, M is the midpoint of CD. BM intersects AC at L and AD produced at E.
To prove: EL = 2BL
Proof: In Δ BMC and Δ EDM

DME = BMC (Vetically opposite angles)
DM = MC (given)

DEM = MBC (alternate angles)
Δ BMC
Δ EDM (ASA congruence)
DE = BC (c.p.c.t)
       
But BC = AD (opposite sides of parallelogram ABCD)
AD = DE
AE = 2AD = 2BC

In Δ AEL and Δ CBL

ALE = BLC (Vertically opposite angles)
AEL = LBC (alternate angles)
Δ AEL ~ Δ CBL (AA similarity axiom)
     

∴ EL = 2 BL
 

Question-17

The side BC of a triangle ABC is bisected at D; O is any point in AD. BO, CO produced meet AC, AB in E,F respectively, and AD is produced to X so that D is the mid point of OX. Prove that AO : AX = AF : AB and show that EF is parallel to BC.

Solution:

Given : The side BC of a triangle ABC is bisected at D; O is any point in AD. BO , CO produced meet AC, AB in E , F respectively, and AD is produced to X so that D is the mid point of OX.

To Prove : AO : AX = AF : AB and show that EF is parallel to BC.
Construction: Join BX and CX.
Proof: In quadrilateral BOCX, BD = DC and DO = DX (given)
BOCX is a parallelogram (When the diagonals of a quadrilateral bisect each other, then the quad. is a parallelogram) 
BX || CO (Definition of a parallelogram)
or BX || FO.
In Δ ABX, BX || FO(proved).
AO : AX = AF : AB (using B.P.T) -----------(i)
Similarly, AO : AX = AE : AC --------------(ii)
From (i) and (ii),
AF : AB = AE : AC
By corollary to B.P.T, EF is parallel to BC.

Question-18

ABC is a triangle in which BAC = 90° and DEFG is a square, prove that DE2 = BD × EC.

Solution:

Given: ABC is a triangle in which
BAC = 90° and DEFG is a square.
To prove: DE2 = BD × EC.
Proof: In Δ AGF and Δ DBG,
AGF = GBD (corresponding angles)
GAF = BDG (each = 90° )
ΔAGF ΔDBG. ----------------------(i)
Similarly, ΔAFG ΔECF (AA Similarity)----------------(ii)
From (i) and (ii), ΔDBG ΔECF.


EF × DG = BD × EC. ---------------------(iii)

Also DEFG is a square
DE = EF = FG = DG ---------------(iv)
From (iii) and (iv),  DE2 = BD × EC.

Question-19

In fig. DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value x.                         
          

Solution:
Given: ABC is a triangle, DE || BC, AD = x, DB = x – 2, AE = x + 2 and EC = x – 1.

To find: x

 

In
Δ ABC, we have

DE || BC
Therefore [By Thale’s theorem]


AD × EC = AE × DB

x(x - 1) = (x - 2)(x + 2)
x2 – x = x2 – 4
x = 4

   


Question-20

In a Δ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4 cm, AE = 8 cm, DB = x – 4
and EC = 3x – 19, find x.

Solution:
 

Given: In Δ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. AD = 4 cm, AE = 8 cm, DB = x - 4 and EC = 3x - 19. 

To find: x.

In
Δ ABC, we have DE || BC
Therefore [By Thale’s theorem]



4(3x - 19) = 8(x - 4)
12x – 76 = 8x – 32
4x = 44
x = 11

 


Question-21

In a Δ ABC, AD is the bisector of A, meeting side BC at D. If AC = 4.2 cm, DC = 6 cm, BC = 10 cm, find AB.

Solution:
 
Given: In a Δ ABC, AD is the bisector of  A, meeting side BC at D. AC = 4.2 cm, DC = 6 cm and BC = 10 cm.

To find: AB. 


In Δ ABC,
[By internal bisector theorem]
AB = 4 × 4.2/6 = 2.8 cm
∴ AB = 2.8 cm

 


Question-22

In Δ ABC, B = 90° and D is the mid-point of BC. Prove that AC2 = AD2 + 3CD2.

Solution:
Given: In Δ ABC, B = 90° and D is the mid-point of BC.
To Prove: AC2 = AD2 + 3CD2
Proof:
In
Δ ABD,
AD2 = AB2 + BD2
AB2 = AD2 - BD2 ……………..(i)

In
Δ ABC,
AC2 = AB2 + BC2
AB2 = AC2 - BC2 ……………..(ii)

Equating (i) and (ii)
AD2 - BD2 = AC2 - BC2
AD2 - BD2 = AC2 – (BD + DC)2
AD2 - BD2 = AC2 – BD2 - DC2 – 2BD
× DC
AD2 = AC2 - DC2 – 2DC2 (DC = BD)
AD2 = AC2 - 3DC2

Question-23

ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC × CD. Prove that BD = BC.

Solution:
Given: A Δ ABC in which AB = AC. D is a point on AC such that BC2 = AC × CD. 
                                        
To prove: BD = BC
Proof: Since BC2 = AC
× CD
Therefore BC
× BC = AC × CD
AC/BC = BC/CD ……………(i)

Also
ACB = BCD
Since
ΔABC ΔBDC [By SAS Axiom of similar triangles]
AB/AC = BD/BC ……………..(ii)

But AB = AC (Given) ……………(iii)

From (i), (ii) and (iii) we get
BD = BC.

 

Question-24

The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that
DF × EF = FB × FA.

Solution:
                                                                                                                                

Given: The diagonal BD of parallelogram ABCD intersects the segment AE at F, where E is any point on BC.
To prove: DF × EF = FB × FA

Proof: In triangles
 AFD and BFE,
FAD = FEB (Alternate angles)
AFD = BFE (Vertically opposite angles)
Therefore
ΔADF ΔBFE (AA similarity)

Hence DF × EF = FB × FA

 
 

Question-25

ABC is a triangle, right-angled at C and AC = 3 BC. Prove that ABC = 60°.

Solution:
Given: ΔABC is right angled at C and AC = 3BC.
 

To prove: ABC = 60°.
Proof:

Let D be the midpoint of AB. Join CD.

Now, AB2 = BC2 + AC2 = BC2 + (BC)2 = 4BC2
Therefore AB = 2BC.
Now, BD = AB = (2BC) = BC.
But, D being the midpoint of hypotenuse AB, it is equidistant from all the three vertices.
Therefore CD = BD = DA or CD = AB = BC.
Thus, BC = BD = CD,

i.e., ΔBCD is a equilateral triangle.
Hence,
ABC = 60°.

Question-26

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Solution:
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Given: ABCD is a quadrilateral in which AD = BC. P, Q, R and S be the midpoints of AB, AC, CD and BD respectively. 

To prove: PQRS is a rhombus.

Proof: By a theorem, line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

In
Δ ABC, PQ||BC and PQ = BC = DA
In
Δ CDA, RQ||DA and RQ = DA
In Δ BDA, SP||DA and SP = DA
In Δ CDB, SR||BC and SR = BC = DA
Therefore SP || RQ, PQ || SR and PQ = RQ = SP = SR.

Hence PQRS is a rhombus.

 


Question-27

Let ABC be a triangle, right-angled at C. If D is the mid-point of BC, prove that AB2 = 4AD2 3AC2.
                                               

Solution:
Given: ABC be a triangle, right-angled at C and D is the mid-point of BC.
To Prove: AB2 = 4AD2 3AC2.
Proof:
From right triangle ACB, we have,

AB2 = AC2 + BC2
      = AC2 + (2CD)2 = AC2 + 4CD2       [Since BC = 2CD]

      = AC2 + 4(AD2 - AC2)                  [From right
Δ ACD]
  
      = 4AD2 3AC2.

Question-28

In the given figure, points D and E trisect BC and B = 90°. Prove that 8AE2 = 3AC2 + 5AD2.

Solution:
Given: In Δ ABC, points D and E trisect on BC and B = 90°.
To Prove: 8AE2 = 3AC2 + 5AD2.
Proof: Let ABC be the triangle in which h
B = 90°. Let the points D and E trisect BC.
Join AD and AE. Then,
AC2 = AB2 + BC2
3AC2 = 3AB2 + 3BC2             ………..(i)

AD2 = AB2 + BD2
5AD2 = 5AB2 + 5BD2             ..……..(ii)

Therefore 3AC2 + 5AD2 = 8AB2 + 3BC2 + 5BD2

                                 = 8AB2 + 3.+ 5

                                 = 8AB2 +BE2

                                 = 8AB2 + 8BE2
                                       
= 8(AB2 + BE2)

                                 = 8AE2.

Question-29

In fig., ABC is a triangle in which AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic.
                                                      

Solution:
Given: In ΔABC, AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE.
To Prove: Points B, C, E and D are concyclic.
Proof: In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that
ABC + CED = 180° and
ACB + BDE = 180°.
In ΔABC, we have
AB = AC and AD = AE
AB - AD = AC - AE
DB = EC

Thus, we have
AD = AE and DB = EC.

DE || BC                                               [By the converse of Thale's Theorem]

ABC = ADE                                         [Corresponding angles]
ABC + BDE = ADE + BDE                   [adding BDE on both sides]
ABC + BDE = 180° 
ACB + BDE = 180°                                  [Since AB = AC Therefore ABC = ACB]
Again DE||BC
ACB = AED
ACB + CED = AED + CED                   [Adding CED on both sides]
ACB + CED = 180° 
ABC + CED = 180°                                        [Since ABC = ACB]
Therefore BDEC is a cyclic quadrilateral.
Hence, B, C, E and D are concyclic points.

Question-30

 In fig,  B < 90° and segment AD ^ BC, show that b2 = h2 + a2 + x2 - 2ax

 

Solution:
Given: In ΔABC, B < 90° and segment AD BC.

To prove: b2 = h2 + a2 + x2 – 2ax

Proof:
b2 = h2 + (a - x)2
b2 = h2 + a2 + x2 - 2ax.

Question-31

In the given figure, ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given area of triangle CPQ = 20 m2, calculate the area of triangle DCP.

 

Solution:
Given: ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q.
Area of triangle CPQ = 20 m2.
 

To Find: Area of triangle DCP.

Construction: Join DB.


BPQ = DPC  (Vertically opposite angles)
BQP = PDC (alternate angles, BQ || DC, DQ meets them)
Δ BPQ Δ CPD (AA similarity)

BP/CP = ½ (Given)



 Area D CPD = Area D BPQ
and area ΔCPQ = 20 cm2 (Given)

Area
ΔBPQ = 10 cm2
Area
ΔCPD = 40 cm2

 (Proportional to bases BC and PC)

 Area Δ DBC = 40 × 3/2 = 60 cm2.




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