# NCERT Solutions Class 10 Ch- 6 Triangles

## Ex 6.1

Q1. Fill in the blanks using correct word given in the brackets:-

(i) All circles are __________. (congruent, similar)

(ii) All squares are __________. (similar, congruent)

(iii) All __________ triangles are similar. (isosceles, equilateral)
Equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

(b) Proportional

Q2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Solution:

(i) Similar figures :- A pair of equilateral triangles of different side lengths, A pair of two rectangles of differnt lengths and breadths, etc.

(ii) Non-similar figures :- A pair of Triangle and Rhombus with angles not equal to 90

Q3. State whether the following quadrilaterals are similar or not:

Solution: From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

## Ex 6.2

Solution:

#### Hence, AD = 2.4 cm.

Q2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Solution:

#### Hence, EF is parallel to QR.

Q3. In the figure, if LM || CB and L

N || CD, prove that AM/AB = AN/AD

Solution:

#### Hence, proved.

Q4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC

Solution:

#### Hence, proved.

Q5. In the figure, DE||OQ and DF||OR, show that EF||QR.

Solution:

#### EF || QR, in ΔPQR.

6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

#### In ΔOQR, BC || QR.

Q7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:

#### Hence, proved, E is the midpoint of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution:

#### Hence, proved.

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Solution:

#### Hence, proved.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Solution:

## Ex 6.3

1. State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution:

#### Hence, ΔDEF ~ ΔPQR

Q2.  In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Solution:

#### ⇒ ∠OAB = 55°

Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

Solution:

#### Hence, proved.

Q4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Solution:

#### ∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Q5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Solution:

#### ∴ ΔRPQ ~ ΔRTS (AA similarity criterion)

Q6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Solution:

#### ∴ ΔADE ~ ΔABC [SAS similarity criterion]

Q7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

Solution:

#### ΔPDC ~ ΔBEC

Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Solution:

#### ∴ ΔABE ~ ΔCFB (AA similarity criterion)

Q9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

Solution:

#### Hence, CA/PA = BC/MP

Q10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

#### ∴ ΔDCA ~ ΔHGF (AA similarity criterion)

Q11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

#### ∴ ΔABD ~ ΔECF (using AA similarity criterion)

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.

#### ΔABC ~ ΔPQR [SAS similarity criterion]

Q13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD

#### Hence, proved.

Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

#### ∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

Q15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

#### Hence, the height of the tower is 42 m.

Q16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Solution: Given, ΔABC ~ ΔPQR

## Ex 6.4

Q1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

#### ⇒ BC = 11.2 cm

Q2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

#### Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

Q3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Q4. If the areas of two similar triangles are equal, prove that they are congruent.

#### Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

#### Hence, Area(ΔDEF):Area(ΔABC) = 1:4

Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

#### ∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2

Q7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

#### ⇒ area(ΔBQC) = 1/2area(ΔAPC)

Tick the correct answer and justify:

Q8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4

#### Hence, the correct answer is (C).

9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81

## Ex 6.5

Q1.  Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Solution:

#### ∴ PM2 = QM × MR

3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD

Solution:

##### ⇒ AD2 = BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

##### AB2 = 2AC2

Q5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

##### Hence, by Pythagoras theorem ΔABC is right angle triangle.

6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

##### Solution: Given, ABC is an equilateral triangle of side 2a.

AB = AC

Hence, BD = DC [by CPCT]

⇒ AD2  = 4a2 – a2

7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution: Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB+ BC+ CD2 + AD= AC+ BD2

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB2 = AO+ BO…………………….. (i) [By Pythagoras theorem]

Similarly,

DC2 = DO+ CO…………………….. (iii)

BC2 = CO+ BO…………………….. (iv)

Adding equations (i) + (ii) + (iii) + (iv), we get,

AB+ AD+ DC+ BC2  =  2(AO+ BO+ DO+ CO)

= 4AO+ 4BO[Since, AO = CO and BO =DO]

= (2AO)+ (2BO)2 = AC+ BD2

AB+ AD+ DC+ BC2 = AC+ BD2

Hence, proved.

8. In Fig. 6.54, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Solution: Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA2 = OF2 + AF2

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC2 = OE2 + EC2

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2

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

(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)

∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Q9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution: Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC2 = AB2 + BC2

102 = 82 + BC2

BC= 100 – 64

BC= 36

BC = 6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Q10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution: Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC2 = AB2 + BC2

242 = 182 + BC2

BC= 576 – 324

BC= 252

BC = 6√7m

Therefore, the distance from the base is 6√7m.

Q11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
hours?

Solution: Given,

Speed of first aeroplane = 1000 km/hr

Distance covered by first aeroplane flying due north in
hours (OA) = 100 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by second aeroplane flying due west in
hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, by Pythagoras Theorem,

AB2 = AO2 + OB2

⇒ AB2 = (1500)2 + (1800)2

⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

Q12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution: Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP2 = PC2 + AC2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25)m2 = 169 m2

AC = 13m

Therefore, the distance between their tops is 13 m.

Q13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Solution: Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC2 + CE2 = AE2 ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC2 + CD2 = BD2 ………………………………..(ii)

From equations (i) and (ii), we get,

AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE2 = CD2 + CE2

In ΔABC, by Pythagoras theorem, we get

AB2 = AC2 + CB2

Putting the above two values in equation (iii), we get

DE2 + AB2 = AE2 + BD2.

Q14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.

Solution: Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

AB2 = AD2 + BD2 ……………………….(i)

AC2 = AD2 + DC2 ……………………………..(ii)

Subtracting equation (ii) from equation (i), we get

AB2 – AC2 = BD2 – DC2

= 9CD2 – CD2  [Since, BD = 3CD]

= 9CD2 = 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB2 – AC2 = BC2/2

⇒ 2(AB2 – AC2) = BC2

⇒ 2AB2 – 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2.

Q15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.

Solution: Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

⇒ 9 AD2 = 7 AB2

Q16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution: Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB2 = AE2 + BE2

4AE2 = 3a2

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence, proved.

17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°

(B) 60°
(C) 90°

(D) 45°

Solution: Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB2 = 108

AC2 = 144

And, BC2 = 36

AB2 + BC2 = AC2

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

## Ex 6.6

1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR

Solution:

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given, PS is the angle bisector of ∠QPR. Therefore,

∠QPS = ∠SPR………………………………..(i)

As per the constructed figure,

∠SPR=∠PRT(Since, PS||TR)……………(ii)

∠QPS = ∠QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,

∠PRT=∠QTR

Therefore,

PT=PR

In △QTR, by basic proportionality theorem,

QS/SR = QP/PT

Since, PT=TR

Therefore,

QS/SR = PQ/PR

Q2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM2 = DN . MC (ii) DN2 = DM . AN.

Let us join Point D and B.

(i) Given,

BD ⊥AC, DM ⊥ BC and DN ⊥ AB

Now from the figure we have,

DN || CB, DM || AB and ∠B = 90 °

Therefore, DMBN is a rectangle.

So, DN = MB and DM = NB

The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)

In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)

In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)

From equation (i) and (ii), we get

∠1 = ∠3

From equation (i) and (iii), we get

∠2 = ∠4

In ∆DCM and ∆BDM,

∴ ∆DCM ∼ ∆BDM (AA similarity criterion)

BM/DM = DM/MC

DN/DM = DM/MC (BM = DN)

⇒ DM2 = DN × MC

Hence, proved.

(ii) In right triangle DBN,

∠5 + ∠7 = 90° ……………….. (iv)

In right triangle DAN,

∠6 + ∠8 = 90° ………………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,

∠6 = ∠7

From equation (v) and (vi), we get,

∠8 = ∠5

In ∆DNA and ∆BND,

∴ ∆DNA ∼ ∆BND (AA similarity criterion)

AN/DN = DN/NB

⇒ DN2 = AN × NB

⇒ DN2 = AN × DM (Since, NB = DM)

Q3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that

AC2= AB2+ BC2+ 2 BC.BD.

Solution: By applying Pythagoras Theorem in ∆ADB, we get,

AB2 = AD2 + DB2 ……………………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,

AC2 = AD2 + (DB + BC) 2

AC2 = AD2 + DB2 + BC2 + 2DB × BC

From equation (i), we can write,

AC2 = AB2 + BC2 + 2DB × BC

Q4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that

AC2= AB2+ BC2 – 2 BC.BD.

Solution: By applying Pythagoras Theorem in ∆ADB, we get,

We can write it as;

⇒ AD2 = AB2 − DB2 ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,

From equation (i),

AB2 − BD2 + DC2 = AC2

AB2 − BD2 + (BC − BD) 2 = AC2

AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD

AC2= AB2 + BC2 − 2BC × BD

Hence, proved.

5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC2 = AD2 + BC.DM + 2 (BC/2) 2

(ii) AB2 = AD2 – BC.DM + 2 (BC/2) 2

(iii) AC2 + AB2 = 2 AD2 + ½ BC2

Solution:

(i) By applying Pythagoras Theorem in ∆AMD, we get,

AM2 + MD2 = AD2 ………………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,

AM2 + MC2 = AC2

AM2 + (MD + DC) 2 = AC2

(AM2 + MD2 ) + DC2 + 2MD.DC = AC2

From equation(i), we get,

AD2 + DC2 + 2MD.DC = AC2

Since, DC=BC/2, thus, we get,

AD+ (BC/2) 2 + 2MD.(BC/2) 2 = AC2

AD+ (BC/2) 2 + 2MD × BC = AC2

Hence, proved.

(ii) By applying Pythagoras Theorem in ∆ABM, we get;

AB2 = AM2 + MB2

=> (AD2 − DM2 ) + MB2

=> (AD2 − DM2 ) + (BD − MD) 2

=> AD2 − DM2 + BD2 + MD2 − 2BD × MD

=> AD2 + BD2 − 2BD × MD

=> AD+ (BC/2) 2 – 2(BC/2) MD

=> AD+ (BC/2) 2 – BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in ∆ABM, we get,

AM2 + MB2 = AB2 ………………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

AM2 + MC2 = AC2 …………………..… (ii)

Adding both the equations (i) and (ii), we get,

2AM2 + MB2 + MC2 = AB2 + AC2

2AM2 + (BD − DM) 2 + (MD + DC) 2 = AB2 + AC2

2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2

2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2

2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD(-BC/2 + BC/2) 2 = AB2 + AC2

2AD+ BC2/2 = AB2 + AC2

Q6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution:

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

DE2 + EA2 = DA2 ……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

DE2 + EB2 = DB2

DE2 + (EA + AB) 2 = DB2

(DE2 + EA2 ) + AB2 + 2EA × AB = DB2

DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

Again, applying Pythagoras theorem in ∆AFC, we get,

AC2 = AF2 + FC2 = AF2 + (DC − FD) 2

= AF2 + DC2 + FD2 − 2DC × FD

= (AF2 + FD2 ) + DC2 − 2DC × FD AC2

AC2= AD2 + DC2 − 2DC × FD ………………… (iii)

Since ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

∠DEA = ∠AFD (Each 90°)

∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF ……………… (vi)

Adding equations (i) and (iii), we get,

DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2

From equation (iv) and (vi),

BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :

(i) ∆APC ~ ∆ DPB

(ii) AP . PB = CP . DP

Solution: Firstly, let us join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

Therefore,

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

∴ AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴AP. PB = PC. DP

Hence, proved.

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB

(ii) PA . PB = PC . PD.

Solution:

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,

∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

Therefore,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

∴ AP. PB = PC. DP

Q9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

Solutions: In the given figure, let us extend BA to P such that;

AP = AC.

Now join PC.

Given, BD/CD = AB/AC

⇒ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,

∠BAD = ∠APC (Corresponding angles) ……………….. (i)

And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;

AP = AC

⇒ ∠APC = ∠ACP ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

Therefore, AD is the bisector of the angle BAC.

Q10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution:

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the

horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;

AC2 = AB2+ BC2

AB= (1.8 m) 2 + (2.4 m) 2

AB= (3.24 + 5.76) m2

AB2 = 9.00 m2

⟹ AB = √9 m = 3m

Thus, the length of the string out is 3 m.

As its given, she pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m