1.    (i)  similar

(ii) similar

(ii) equilateral

(iv) equal, proportional

2.    (i) Two different examples of pair of similar figures are:

(a) Any two rectangles

(b) Any two squares

(ii) Two different examples of pair of non-similar figures are:

A scalene and an equilateral triangle

An equilateral triangle and a right angled triangle

3.   On looking at the given figures of the quadrilaterals, we can say that they are not similar because their angles are not equal.

4.   

5.   (i) Given: PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm

                 

Therefore, EF does not divide the sides PQ and PR of Δ PQR in the same ratio.

∴ EF is not parallel to QR.

(ii) Given:       PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Therefore, EF divides the sides PQ and PR of Δ PQR in the same ratio.

∴ EF is parallel to QR.

(iii) Given:    PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
⇒ EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And     ER = PR – PF = 2.56 – 0.36 = 2.20 cm

Therefore, EF divides the sides PQ and PR of Δ PQR in the same ratio.

∴ EF is parallel to QR.

6.   In Δ ABC,       LM || CB

7.  In Δ BCA,  DE || AC

 

8.   In Δ PQO,  DE || OQ

9.   Given: 0 is any point in Δ PQR, in which AB || PQ and AC || PR.

To prove:       BC || QR

Construction: Join BC.

Proof: In Δ OPQ,        AB || PQ

∴ In Δ OQR, B and C are points dividing the sides OQ and OR in the same ratio.

∵ By the converse of Basic Proportionality theorem,

⇒ BC || QR

10.   Given: A triangle ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC at E.

To prove:         AE = EC

Proof: Since DE || BC

Hence, E is the mid-point of the third side AC.

11.   Given: A triangle ABC, in which D and E are the mid-points of sides AB and AC respectively.

To Prove: DE || BC

Proof: Since D and E are the mid-points of AB and AC respectively.

AD = DB and AE = EC

Now, AD = DB

Thus, in triangle ABC, D and E are points dividing the sides AB and AC in the same ratio. Therefore, by the converse of Basic Proportionality theorem, we have DE || BC

12.  Given: A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at 0.

To Prove: 

Construction: Through 0, draw OE || AB, i.e. OE || DC.

Proof: In Δ ADC, we have OE || DC

13.  Given: A quadrilateral ABCD, in which its diagonals AC and BD intersect each other at 0 such that i.e. 

To Prove: Quadrilateral ABCD is a trapezium.

Construction: Through 0, draw OE || AB meeting AD at E.

Proof: In Δ ADB, we have OE || AB                  [By construction]

Thus in Δ ADC, E and 0 are points dividing the sides AD and AC in the same ratio. Therefore by the converse of Basic Proportionality theorem, we have

EO || DC

But EO || AB                                              [By construction]

∴ AB || DC

∴ Quadrilateral ABCD is a trapezium.

14.   (i) In Δs ABC and PQR, we observe that,

LA = ZP = 600, LB = LQ = 800 and LC = ZR = 400

∴ By AAA criterion of similarity, ΔABC – ΔPQR

(ii) In Δs ABC and PQR, we observe that,

∴ By SSS criterion of similarity,  ΔABC – ΔPQR

(iii) In Δs LMP and DEF, we observe that, the ratio of the sides of these triangles is not equal. Therefore, these two triangles are not similar.

(iv) In Δs MNL and QPR, we observe that, ∠M = ∠Q = 700

∴ These two triangles are not similar as they do not satisfy SAS criterion of similarity.

(v) In Δs ABC and FDE, we have,         ∠A = ∠F = 800

 

∴ These two triangles are not similar as they do not satisfy SAS criterion of similarity.

(vi) In Δs DEF and PQR, we have, ∠D = ∠P = 70°             [∵ ∠P = 1800 – 800 – 300 = 700]
And ∠E = ∠Q = 80°

By AAA criterion of similarity, ΔDEF – ΔPQR

15.   Since BD is a line and OC is a ray on it.

∴ ∠DOC + ∠BOC = 1800                 ⇒           ∠DOC + 125° = 1800

⇒ ∠DOC = 550

In Δ CDO, we have ∠CDO + ∠DOC + ∠DCO = 180°

⇒ 700 + 550 + ∠DCO = 1800          ⇒           ∠DCO = 550

It is given that Δ ODC – Δ OBA

∴ ∠OBA = ∠ODC, ∠OAB = ∠OCD       ⇒         ∠OBA = 70°, ∠OAB = 55°

Hence ∠DOC = 550, ∠DCO = 550 and ∠OAB = 550

16.  Given: ABCD is a trapezium in which AB || DC.

To Prove: 

Proof: In Δs OAB and OCD, we have,

  ∠ 5 = ∠ 6         [Vertically opposite angles]

  ∠ 1 = ∠ 2          [Alternate angles]

And  ∠ 3 = ∠ 4         [Alternate angles]

∴ By AAA criterion of similarity,  ΔOAB – ΔODC

Hence, 

17.  We have   

Also,       ∠1 = ∠2         [Given]

PR = PQ          (2)        [∵ Sides opposite to equal ∠s are equal]

From eq. (1) and (2), we get 

In Δs PQS and TQR, we have 

∴ By SAS criterion of similarity, Δ PQS – Δ TQR

18.   In Δs RPQ and RTS, we have

∠ RPQ = ∠ RTS           [Given]

∠ PRQ = ∠ TRS           [Common]

∴ By AA-criterion of similarity,

Δ RPQ – RTS

19.  It is given that Δ ABE = Δ ACD

And         ∠ BAC = ∠ DAE          [Common]

Thus, by SAS criterion of similarity,    Δ ADE – Δ ABC

20.  (i) In Δs AEP and CDP, we have,

AEP = ∠ CDP = 900                              [∵ CE ⊥ AB, AD ⊥ BC]

And ∠ APE =  CPD                             [Vertically opposite]

∴ By AA-criterion of similarity,             Δ AEP – Δ CDP

(ii) In Δs ABD and CBE, we have,

∠ ADB = ∠ CEB = 900

And ∠ ABD = ∠ CBE                [Common]

∴ By AA-criterion of similarity,            Δ ABD — Δ CBE

(iii) In Δs AEP and ADB, we have,

∠ AEP = ∠ ADB = 900                    [∵ AD ⊥ BC, CE ⊥ AB]

And  ∠ PAE = ∠ DAB                      [Common]

∴ By AA-criterion of similarity,         Δ AEP – Δ ADB

(iv) In Δs PDC and BEC, we have,

∠ PDC = ∠ BEC = 900                     [∵ CE ⊥ AB, AD ⊥ BC]

And ∠ PCD = ∠ BEC                       [Common]

∴ By AA-criterion of similarity,     ΔPDC – ΔBEC

21.  In Δs ABE and CFB, we have,

∠ AEB = ∠ CBF                       [Alt. ∠ s]

∠ A = ∠ C                                 [opp. ∠ s of a || gm]

∴ By AA-criterion of similarity, we have Δ ABE — Δ CFB

22.  (i) In Δs ABC and AMP, we have,

∠ ABC = ∠ AMP = 900            [Given]

∠ BAC = ∠ MAP                      [Common angles]

∴ By AA-criterion of similarity, we have

Δ ABC — Δ AMP

(ii) We have         Δ ABC — Δ AMP        [As prove above]

23.  We have,     Δ ABC ∼ Δ FEG

[∵ CD and GH are bisectors of ∠ C and ∠ G respectively]

In Δs DCA and HGF, we have

∠ A = ∠ F [From eq.(1)]

∠ 2 = ∠ 4   [From eq.(2)]

∴ By AA-criterion of similarity, we have

Δ DCA — Δ HGF

Which proves the (iii) part

We have,         Δ DCA — Δ HGF

In Δs DCA and HGF, we have

∠ 1 = ∠ 3           [From eq.(2)]

∠ B = ∠ E            [∵ ΔDCB — Δ HE]

Which proves the (ii) part

24.  Here ΔABC is isosceles with AB = AC

∴ ∠B = ∠C

In Δs ABD and ECF, we have

∠ ABD = ∠ ECF                        [∵ ∠B = ∠C]

∠ ABD = ∠ ECF = 900              [∵ AD ⊥ BC and EF ⊥ AC]

∴ By AA-criterion of similarity, we have Δ ABD – Δ ECF

25.  Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that 

To Prove: Δ ABC — Δ PQR

Proof:

∴ Δ ABD – Δ PQM            [By SSS-criterion of similarity]

∠ B = ∠ Q        [Similar triangles have corresponding angles equal]

And                          [Given]

∴ By SAS-criterion of similarity, we have

Δ ABC — Δ PQR

26.  In triangles ABC and DAC,

∠ ADC = ∠ BAC                      [Given]

and ∠ C = ∠ C                            [Common]

By AA-similarity criterion,

 Δ ABC — Δ DAC

27.  Given: AD is the median of Δ ABC and PM is the median of  PQR such that

To Prove:  Δ ABC — Δ PQR

Proof:

  

∴ Δ ABD – Δ PQM                       [By SSS-criterion of similarity]

 ⇒ ∠ B = ∠ Q       [Similar triangles have corresponding angles equal]

And                        [Given]

∴ By SAS-criterion of similarity, we have

Δ ABC — Δ PQR

28.  Let AB the vertical pole and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Joined BC and EF.

        

Let DE = x meters

Here, AB = 6 m, AC = 4 m and DF = 28 m

In the triangles ABC and DEF,

∠A = ∠D= 900

And     ∠ C = ∠ F        [Each is the angular elevation of the sun]

∴ By AA-similarity criterion,

Δ ABC — Δ DEF

Hence, the height of the tower is 42 meters.

29.  Given: AD and PM are the medians of triangles ABC and PQR respectively, where Δ ABC — Δ PQR

To prove: 

Proof: In triangles ABD and PQM,

∠B = ∠Q [Given]

And     [∵ AD and PM are the medians of BC and QR respectively]

 

30.   We have,

31.   In Δs AOB and COD, we have,

∠ AOB = ∠ COD         [Vertically opposite angles]

∠ OAB = ∠ OCD         [Alternate angles]

 

By AA-criterion of similarity,

 

32.   Given: Two Δs ABC and DBC which stand on the same base but on the opposite sides of BC.

                    

Construction : Draw AE ⊥ BC and DF ⊥ BC.

33.   Given: Two Δs ABC and DEF such that Δ ABC – Δ DEF And Area (ΔABC) = Area (A DEF)

To Prove: Δ ABC ≅ Δ DEF

Proof:  Δ ABC – Δ DEF

To establish Δ ABC = Δ DEF, it is sufficient to prove that, AB = DE, BC = EF and AC = DF Now, Area (Δ ABC) = Area (Δ DEF)

34.  Since, D and E are the mid-points of the sides BC and CA of Δ ABC respectively.

∴ DE || BA             ⇒              DE || FA           …………. (i)

Since, D and F are the mid-points of the sides BC and AB of Δ ABC respectively.

∴ DF || CA             ⇒               DE || AE         …………. (ii)
From (i) and (ii), we can say that AFDE is a parallelogram. Similarly, BDEF is a parallelogram.

Now, in Δs DEF and ABC, we have

∠ FDE = ∠ A             [opposite angles of || gm AFDE]

And ∠ DEF = ∠ B            [opposite angles of || gm BDEF]

∴ By AA-criterion of similarity, we have Δ DEF — Δ ABC

Hence, Area (Δ DEF) : Area (Δ ABC) = 1 : 4

35.  Given: Δ ABC — Δ PQR, AD and PM are the medians of Δs ABC and PQR respectively.

36.  Given: A square ABCD, Equilateral Δs BCE and ACF have been drawn on side BC and the diagonal AC respectively.

To Prove: Area (Δ BCE) =   Area (Δ ACF)

Proof: Δ BCE – Δ ACF [Being equilateral so similar by AAA criterion of similarity]

37. (C) Since Δ ABC and Δ BDE are equilateral, they are equiangular and hence,

ΔABC – ΔBDE

38.  (D) Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Therefore,

39.  (i) Let a= 7 cm, b = 24 cm and c = 25 cm Here the larger side is c = 25 cm.

We have, a2 + b2 = 72 + 242 = 49 +576 = 625 = c2

So, the triangle with the given sides is a right triangle. Its hypotenuse = 25 cm

(ii) Let a= 3 cm, b = 8 cm and c = 6 cm

Here the larger side is b = 8 cm.

We have, a2 + c2 = 32 + 62 = 9 + 36 = 45 ≠ b2

So, the triangle with the given sides is not a right triangle.

(iii) Let a= 50 cm, b = 80 cm and c = 100 cm Here the larger side is c = 100 cm.

We have, a2 + b2 = 502 + 802 = 2500 + 6400 = 8900 ≠ c2

So, the triangle with the given sides is not a right triangle.

(iv) Let a= 13 cm, b = 12 cm and c = 5 cm Here the larger side is a= 13 cm.

We have, b2 + c2 =122+52 =144 + 25 =169 = a2

So, the triangle with the given sides is a right triangle. Its hypotenuse = 13 cm

40.  Given: PQR is a triangle right angles at P and PM ⊥ QR

To Prove: PM2 = QM.MR

Proof: Since PM ⊥ QR

41.  Given: ABD is a triangle right angled at A and AC ⊥ BD.

To Prove: (i) AB2 = BC.BD,      (ii) AC2 = BC.DC,       (iii) AD2 = BD.CD

Proof: (i) Since AC ⊥ BD

∴ Δ ABC – Δ ADC and each triangle is similar to Δ ABD

∴ Δ ABC – Δ ABD

42.  Since ABC is an isosceles right triangle, right angled at C.

⇒ AB2 = AC2 + BC2

⇒ AB2 = AC2 + AC2                              [BC = AC, given]

⇒ AB2 = 2AC2

43.  Since ABC is an isosceles right triangle with AC = BC and AB2 = 2AC2

⇒ AB2 = AC2 + AC2

⇒ AB2 = AC2 + BC2                               [BC = AC, given]

∴ Δ ABC is right angled at C.

44.  Let ABC be an equilateral triangle of side 2a units.

Draw AD ⊥ BC. Then, D is the mid-point of BC.

Since, ABD is a right triangle, right triangle at D.

∴ AB2 = AD2 + BD2            ⇒               (2a)2= AD2 + (a)2
⇒ AD2 = 4a2 – a2 = 3a2

Each of its altitude = √3a

45.  Let the diagonals AC and BD of rhombus ABCD intersect each other at 0. Since the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 900 and AO= CO, BO = OD

Since AOB is a right triangle, right angled at 0.

∴ AB2 = 0A2 + OB2

⇒                                          4AB2 = AC2 + BD2       ……… (1)

Similarly, we have             4BC2 = AC2 + BD2      ………. (2)

4CD2 = AC2 + BD2     ……….. (3)

4AD2 = AC2 + BD2      ………. (4)

Adding all these results, we get

     4(AB2 + BC2 + CD2 + DA2) = 4(AC2 + BD2)

⇒ AB2 + BC2 + CD2 + DA2 = AC2 + BD2

46.   Join AO, BO and CO.

(i) In right A s OFA, 0DB and OEC, we have

OA2 = AF2 + OF2,          OB2 = BD2 + OD2        and OC2 = CE2 + OE2

Adding all these, we get

OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2

⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

(ii) In right A s 0DB and ODC, we have

OB2 = BD2 + OD2                 and 0C2 = OD2 + CD2

⇒ OB2 – 0C2 = BD2 – CD2 ………………….. (1)

Similarly, we have       OB2 – 0C2 = BD2 – CD2 (2)

and      OB2 – 0C2 = BD2 – CD2             ………… (3)

Adding equations (1), (2) and (3), we get

(OB2 – OC2) + (OC2 – OA2) + (OA2 – OB2) = (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2)

⇒ (BD2 + CE2 + AF2) – (AE2 + CD2 + BF2) = 0

⇒ AF2 + BD2 + CE2 = AE2 + BF2 + CD2

47.  Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a right triangle, right angled at C.

⇒ AB2 = AC2 + BC2                ⇒          102 = AC2 + 82

⇒ AC2 = 100 – 64                   ⇒           AC2 = 36

⇒ AC = 6

Hence, the foot of the ladder is at a distance 6 m from the base of the wall.

48.  Let AB (= 24m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.

⇒ AB2 = AC2 + BC2          ⇒           242 = AC2 + 182

⇒ AC2 = 576 – 324            ⇒           AC2 = 252

⇒ AC = 6 √7

Hence, the stake may be placed at distance of 6√7 m from the base of the pole.

49.  Let the first aeroplane starts from O and goes upto A towards north where

                       

Let the second aeroplane starts from 0 at the same time and goes upto B towards west where

According to the question the required distance = BA

In right angled triangle ABC, by Pythagoras theorem, we have,

AB2 = 0A2 + 0B2 = (1500)2 + (1800)2 = 2250000 + 3240000 = 5490000 = 9 × 61 × 100 × 100

⇒ AB = 300√61  km

50. Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m

Draw CE ⊥ AB and join AC.

∴ CE = DB = 12 m

AE = AB – BE = AB – CD = (11 – 6)m = 5 m

In right angled triangle ACE, by Pythagoras theorem, we have

AC2 = CE2 + AE2 = 122 + 52 = 144 + 25 = 169

⇒ AC = 13 m

Hence, the distance between the tops of the two poles is 13 m.

51.   In right angled Δs ACE and DCB, we have

AE2 = AC2 + CE2 and BD2 = DC2 + BC2

⇒ AE2 + BD2 = (AC2 + CE2) + (DC2 + BC2)

⇒ AE2 + BD2 = (AC2 + BC2) + (DC2 + CE2)

⇒ AE2 + BD2 = AB2 + DE2

[By Pythagoras theorem, AC2 + BC2 = AB2 and DC2 + CE2 = DE2]

52.  We have,      DB = 3CD

Since, Δ ABD is a right triangle, right angled at D. Therefore by Pythagoras theorem, we have,     AB2 = AD2 + DB2              …. (2)

Similarly, from Δ ACD, we have,        AC2 = AD2 + CD2    …. (3)

From eq. (2) and (3)    AB2 – AC2 = DB2 – CD2

53.  Let ABC be an equilateral triangle and let D be a point on BC such that BD =  BC

Draw AE ⊥ BC, Join AD.

In Δs AEB and AEC, we have,

AB = AC                             [∵ A ABC is equilateral]

∠ AEB = ∠ AEC                [∵ each 900]

And         AE = AE

∴ By SAS-criterion of similarity, we have

Δ AEB – Δ AEC

⇒ BE = EC

Thus, we have,  

Since, ∠C = 600

∴ Δ ADC is an acute angle triangle.

54.   Let ABC be an equilateral triangle and let AD ⊥ BC. In Δs ADB and ADC, we have,

AB = AC                          [Given]

∠B = ∠C = 600              [Given]

And ∠ ADB = ∠ ADC   [ Each = 900]

ΔADB  ≅ ΔADC            [ By RHS criterion of congruence]

Since ΔADB is a right triangle, right angled at D, by Pythagoras theorem, we have,

55.  (C) In ΔABC, we have, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

Now, AB2 + BC2 = (6√3)2 + (6)2 = 36 × 3 +36 =108+36= 144 = (AC)2

Thus, ΔABC is a right triangle, right angled at B.

∴ ∠ B = 900

Let D be the mid-point of AC. We know that the mid-point of the hypotenuse of a right triangle is equidistant from the vertices.

Also, BC = 6 cm

∴ In ΔBDC, we have, BD = CD = BC

⇒ Δ BDC is equilateral             ⇒                 ∠ ACB = 600

∴ ∠A = 1800 – (∠B + ∠C) = 1800 — (900 + 600) = 300

Thus, ∠A = 300 and ∠B = 900

56.  Given: PQR is a triangle and PS is the internal bisector of Z QPR meeting QR at S.

∴ ∠ QPS = ∠ SPR

To Prove: 

Construction: Draw RT || SP to cut QP produced at T.

Proof: Since PS || TR and PR cuts them, hence,

        ∠ SPR = ∠ PRT           …(i)           [Alternate ∠ s]

And ∠ QPS = ∠ PTR          …(ii)          [Corresponding ∠ s]

But   ∠ QPS = ∠ SPR                             [Given]

          ∠ PRT = ∠ PTR                            [From eq. (i) & (ii)]

⇒       PT = PR          …(iii)                    [Sides opposite to equal angles are equal]

Now, in Δ QRT,

RT || SP          [By construction]

57.  Since, AB ⊥ BC and DM ⊥ BC

⇒ AB || DM

Similarly, BC ⊥ AB and DN ⊥ AB

⇒ CB || DN

∴  quadrilateral BMDN is a rectangle.

∴   BM = ND

(i) In Δ BMD,    ∠ 1 + ∠ BMD + ∠ 2 = 1800

⇒ ∠ 1 + 900 + ∠ 2 = 1800           ⇒         ∠ 1 + ∠ 2 = 900

Similarly in Δ DMC,                                ∠ 3+ ∠ 4= 900

Since       BD ⊥ AC,

∴ ∠ 2 + ∠ 3 = 900

Now, ∠ 1 + ∠ 2 = 900 and ∠ 2 + ∠ 3 = 900

⇒ ∠ 1 + ∠ 2 = ∠ 2 + ∠ 3            ⇒             ∠ 1 = ∠ 3

Also,            ∠ 3 + ∠ 4 = 900 and ∠ 2 + ∠ 3 = 900

⇒ ∠ 3 + ∠ 4 = ∠ 2 + ∠ 3            ⇒             ∠ 4 = ∠ 2

Thus, in Δ BMD and Δ DMC,

∠ 1 = ∠ 3 and ∠ 4 = ∠ 2

∴ Δ BMD – Δ DMC

(ii) Processing as in (i), we can prove that

Δ BND – Δ DNA

58.  Given: ABC is a triangle in which ∠ ABC > 900 and AD ⊥ CB produced.

To prove: AC2 = AB2 + BC2 + 2BC.BD

Proof: Since Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AB2 = AD2 + DB2             …(i)

Again, Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

AC2 = AD2 + DC2

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB.BC

AC2 = (AD2 + DB2)+ BC2 + 2DB.BC

AC2 = AB2 + BC2 + 2DB.BC

[Using eq. (i)]

59.  Given: ABC is a triangle in which ∠ ABC < 900 and AD ⊥ BC

To prove: AC2 = AB2 + BC2 – 2BC.BD

Proof: Since Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AB2 = AD2 + BD2              … (i)

Again, Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem,

AC2 = AD2 + DC2

AC2 = AD2 + (BC – BD)2

AC2 = AD2 + BC2 + BD2 – 2BC.BD

AC2 = (AD2 + DB2)+ BC2 – 2DB.BC

AC2 = AB2 + BC2 – 2DB.BC

[Using eq. (i)]

60. Since ∠ AMD = 900, therefore ∠ ADM < 900 and ∠ ADC > 900 Thus, ∠ ADC is acute angle and ∠ ADC is obtuse angle.

(i) In Δ ADC, ∠ ADC is an obtuse angle.

∴  AC2 = AD2 + DC2 + 2DC.DM

(ii) In Δ ABD, ∠ ADM is an acute angle.

AB2 = AD2 + BD2 – 2BD.DM

(iii) From eq. (i) and eq. (ii),

61.  If AD is a median of Δ ABC, then

             [See Q.60 (iii)]

Since the diagonals of a parallelogram bisect each other, therefore, BO and DO are medians of triangles ABC and ADC respectively.

Adding eq. (i) and (ii),

AB2 + BC2 + AD2 + CD2 = 2 (B02 + D02) + AC2

62.  (i) In the triangles APC and DPB,

∠ APC = ∠ DPB        [Vertically opposite angles]

∠ CAP = ∠ BDP        [Angles in same segment of a circle are equal]

∴ By AA-criterion of similarity,

Δ APC – Δ DPB

(ii) Since    Δ APC – Δ DPB

63.  (i) In the triangles PAC and PDB,

∠ APC = ∠ DPB  [Common]

∠ CAP = ∠ BDP            [∵ ∠ BAC = 1800 – ∠ PAC and ∠ PDB = CDB] = 1800 – ∠ BAC = 1800 – (1800 – ∠ PAC) = ∠ PAC]

∴ By AA-criterion of similarity,

Δ APC – Δ DPB

(ii) Since      Δ APC – Δ DPB

64.  Given: ABC is a triangle and D is a point on BC such that 

To prove: AD is the internal bisector of ∠ BAC.

Construction: Produce BA to E such that AE = AC. Join CE.

Proof: In Δ AEC,         since AE = AC

∠ AEC = ∠ ACE ………………. (i)

[Angles opposite to equal side of a triangle are equal]

∴ By converse of Basic Proportionality Theorem,

DA || CE

Now,        since CA is a transversal,

∴              ∠ BAD = ∠ AEC        … (ii)          [Corresponding ∠ s]

And         ∠ DAC = ∠ ACE      …  (iii)        [Alternate ∠ s]

Also         ∠ AEC = ∠ ACE          [From eq. (i)]

Hence,    ∠ BAD = ∠ DAC          [From eq. (ii) and (iii)]

Thus, AD bisects ∠ BAC internally.

65.  I. To find: The length of AC.

By Pythagoras theorem,

AC2 = (2.4)2 + (1.8)2

⇒ AC2 = 5.76 + 3.24 = 9.00

⇒ AC = 3 m

∴ Length of string she has out= 3 m

Length of the string pulled at the rate of 5 cm/sec in 12 seconds

= (5 x 12) cm = 60 cm = 0.60 m

∴ Remaining string left out = 3 – 0.6 = 2.4 m

II. To find: The length of PB

PB2 = PC2 – BC2

= (2.4)2 – (1.8)2

= 5.76 – 3.24 = 2.52

Hence, the horizontal distance of the fly from Nazima after 12 seconds

= 1.59 + 1.2 = 2.79 m (approx.)