**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.

**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.

**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 = 60^{0}, LB = LQ = 80^{0} and LC = ZR = 40^{0}

∴ 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 = 70^{0}

∴ 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 = 80^{0}

∴ 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 = 180^{0} – 80^{0} – 30^{0} = 70^{0}]

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 = 180^{0} ⇒ ∠DOC + 125° = 180^{0}

⇒ ∠DOC = 55^{0}

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

⇒ 70^{0} + 55^{0} + ∠DCO = 180^{0} ⇒ ∠DCO = 55^{0}

It is given that Δ ODC – Δ OBA

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

Hence ∠DOC = 55^{0}, ∠DCO = 55^{0} and ∠OAB = 55^{0}

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

**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

Also, ∠1 = ∠2 [Given]

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

∴ 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 = 90^{0} [∵ 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 = 90^{0}

And ∠ ABD = ∠ CBE [Common]

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

(iii) In Δs AEP and ADB, we have,

∠ AEP = ∠ ADB = 90^{0} [∵ 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 = 90^{0} [∵ 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 = 90^{0} [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 = 90^{0} [∵ 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]

∴ 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

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

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

∴ 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= 90^{0}

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

**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, *a ^{2} + b^{2} = *7

^{2}+ 24

^{2}= 49 +576 = 625

*= c*

^{2}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, *a ^{2} + c^{2} = *3

^{2}

^{ }+ 6

^{2 }= 9 + 36 = 45 ≠

*b*

^{2}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, *a ^{2} + b^{2} = *50

^{2}+ 80

^{2}= 2500 + 6400 = 8900 ≠

*c*

^{2}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, b^{2} + c^{2} =12^{2}+5^{2} =144 + 25 =169 = *a ^{2}*

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: **PM^{2}** = **QM.MR

**Proof: **Since PM ⊥ QR

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

**To Prove: **(i) AB^{2} = BC.BD, (ii) AC^{2} = BC.DC, (iii) AD^{2} = 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.

⇒ AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = AC^{2} + AC^{2} [BC = AC, given]

⇒ AB^{2} = 2AC^{2}

**43.** Since ABC is an isosceles right triangle with AC = BC and AB^{2} = 2AC^{2}

⇒ AB^{2} = AC^{2} + AC^{2}

⇒ AB^{2} = AC^{2} + BC^{2} [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.

∴ AB^{2} = AD^{2} + BD^{2} ⇒ (2a)^{2}= AD^{2} + (a)^{2}

⇒ AD^{2} = 4a^{2} – a^{2} = 3a^{2}

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 = 90^{0} and AO= CO, BO = OD

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

∴ AB^{2} = 0A^{2} + OB^{2}

⇒ 4AB^{2} = AC^{2} + BD^{2} ……… (1)

Similarly, we have 4BC^{2} = AC^{2} + BD^{2} ………. (2)

4CD^{2} = AC^{2} + BD^{2} ……….. (3)

4AD^{2} = AC^{2} + BD^{2} ………. (4)

Adding all these results, we get

4(AB^{2} + BC^{2} + CD^{2} + DA^{2}) = 4(AC^{2} + BD^{2})

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

**46.** Join AO, BO and CO.

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

OA^{2} = AF^{2} + OF^{2}, OB^{2} = BD^{2} + OD^{2} and OC^{2} = CE^{2} + OE^{2}

Adding all these, we get

OA^{2} + OB^{2} + OC^{2} = AF^{2} + BD^{2} + CE^{2} + OF^{2} + OD^{2} + OE^{2}

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

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

OB^{2} = BD^{2} + OD^{2} and 0C^{2} = OD^{2} + CD^{2}

⇒ OB^{2} – 0C^{2} = BD^{2} – CD^{2} ………………….. (1)

Similarly, we have OB^{2} – 0C^{2} = BD^{2} – CD^{2} (2)

and OB^{2} – 0C^{2} = BD^{2} – CD^{2} ………… (3)

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

(OB^{2} – OC^{2}) + (OC^{2} – OA^{2}) + (OA^{2} – OB^{2}) = (BD^{2} – CD^{2}) + (CE^{2} – AE^{2}) + (AF^{2} – BF^{2})

⇒ (BD^{2} + CE^{2} + AF^{2}) – (AE^{2} + CD^{2} + BF^{2}) = 0

⇒ AF^{2} + BD^{2} + CE^{2} = AE^{2} + BF^{2} + CD^{2}

**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.

⇒ AB^{2} = AC^{2} + BC^{2} ⇒ 10^{2} = AC^{2} + 8^{2}

⇒ AC^{2} = 100 – 64 ⇒ AC^{2} = 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.

⇒ AB^{2} = AC^{2} + BC^{2} ⇒ 24^{2} = AC^{2} + 18^{2}

⇒ AC^{2} = 576 – 324 ⇒ AC^{2} = 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,

AB^{2} = 0A^{2} + 0B^{2} = (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

AC^{2} = CE^{2} + AE^{2} = 12^{2} + 5^{2} = 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

AE^{2} = AC^{2} + CE^{2} and BD^{2} = DC^{2} + BC^{2}

⇒ AE^{2} + BD^{2} = (AC^{2} + CE^{2}) + (DC^{2} + BC^{2})

⇒ AE^{2} + BD^{2} = (AC^{2} + BC^{2}) + (DC^{2} + CE^{2})

⇒ AE^{2} + BD^{2} = AB^{2} + DE^{2}

[By Pythagoras theorem, AC^{2} + BC^{2} = AB^{2} and DC^{2} + CE^{2} = DE^{2}]

**52.** We have, DB = 3CD

Since, Δ ABD is a right triangle, right angled at D. Therefore by Pythagoras theorem, we have, AB^{2} = AD^{2} + DB^{2} …. (2)

Similarly, from Δ ACD, we have, AC^{2} = AD^{2} + CD^{2} …. (3)

From eq. (2) and (3) AB^{2} – AC^{2} = DB^{2} – CD^{2}

**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 90^{0}]

And AE = AE

∴ By SAS-criterion of similarity, we have

Δ AEB – Δ AEC

⇒ BE = EC

Since, ∠C = 60^{0}

∴ 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 = 60^{0} [Given]

And ∠ ADB = ∠ ADC [ Each = 90^{0}]

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, AB^{2} + BC^{2} = (6√3)^{2 }+ (6)^{2} = 36 × 3 +36 =108+36= 144 = (AC)^{2}

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

∴ ∠ B = 90^{0}

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 = 60^{0}

∴ ∠A = 180^{0} – (∠B + ∠C) = 180^{0} — (90^{0} + 60^{0}) = 30^{0}

Thus, ∠A = 30^{0} and ∠B = 90^{0}

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

∴ ∠ QPS = ∠ SPR

**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 = 180^{0}

⇒ ∠ 1 + 90^{0} + ∠ 2 = 180^{0} ⇒ ∠ 1 + ∠ 2 = 90^{0}

Similarly in Δ DMC, ∠ 3+ ∠ 4= 90^{0}

Since BD ⊥ AC,

∴ ∠ 2 + ∠ 3 = 90^{0}

Now, ∠ 1 + ∠ 2 = 90^{0} and ∠ 2 + ∠ 3 = 90^{0}

⇒ ∠ 1 + ∠ 2 = ∠ 2 + ∠ 3 ⇒ ∠ 1 = ∠ 3

Also, ∠ 3 + ∠ 4 = 90^{0} and ∠ 2 + ∠ 3 = 90^{0}

⇒ ∠ 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 > 90^{0} and AD ⊥ CB produced.

**To prove: **AC^{2} = AB^{2} + BC^{2} + 2BC.BD

**Proof: **Since Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AB^{2} = AD^{2} + DB^{2} …(i)

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

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AD^{2} + (DB + BC)^{2}

AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB.BC

AC^{2} = (AD^{2} + DB^{2})+ BC^{2} + 2DB.BC

AC^{2} = AB^{2} + BC^{2} + 2DB.BC

[Using eq. (i)]

**59.** **Given: **ABC is a triangle in which ∠ ABC < 90^{0} and AD ⊥ BC

**To prove: **AC^{2} = AB^{2} + BC^{2} – 2BC.BD

**Proof: **Since Δ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, AB^{2} = AD^{2} + BD^{2} … (i)

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

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AD^{2} + (BC – BD)^{2}

AC^{2} = AD^{2} + BC^{2} + BD^{2} – 2BC.BD

AC^{2} = (AD^{2} + DB^{2})+ BC^{2} – 2DB.BC

AC^{2} = AB^{2} + BC^{2} – 2DB.BC

[Using eq. (i)]

**60.** Since ∠ AMD = 90^{0}, therefore ∠ ADM < 90^{0} and ∠ ADC > 90^{0} Thus, ∠ ADC is acute angle and ∠ ADC is obtuse angle.

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

∴ AC^{2} = AD^{2} + DC^{2} + 2DC.DM

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

AB^{2} = AD^{2} + BD^{2} – 2BD.DM

(iii) From eq. (i) and eq. (ii),

**61.** If AD is a median of Δ ABC, then

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),

AB^{2} + BC^{2} + AD^{2} + CD^{2} = 2 (B0^{2} + D0^{2}) + AC^{2}

**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 = 180^{0} – ∠ PAC and ∠ PDB = CDB] = 180^{0} – ∠ BAC = 180^{0} – (180^{0} – ∠ 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,

AC^{2} = (2.4)^{2} + (1.8)^{2}

⇒ AC^{2} = 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

PB^{2} = PC^{2} – BC^{2}

= (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.)