Question 1: D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC such that DE || BC.

(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.

(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.

(iii) If \(\frac{AD}{DB}=\frac{4}{7}\) and AC = 6.6 cm, find AE.

(iv) If \(\frac{AD}{AB}=\frac{8}{15}\) and EC = 3.5 cm, find AE.

Solution:

(i) In \(\Delta\)ABC, DE || BC, AD = 3.6 cm, AB = 10 cm, AE = 4.5 cm

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)

\(\Rightarrow\) \(\frac{3.6}{AB-AD}=\frac{4.5}{AC-AE}\)

\(\Rightarrow\) \(\frac{3.6}{10-3.6}=\frac{4.5}{AC-4.5}\)

\(\Rightarrow\) \(\frac{3.6}{6.4}=\frac{4.5}{AC-4.5}\)

\(\Rightarrow\) 3.6AC – 16.2 = 28.8

\(\Rightarrow\) 3.6AC = 45

\(\Rightarrow\) AC = 12.5 cm

Therefore, EC = AC – AE = 12.5 – 4.5 = 8 cm

Hence, AC = 12.5 cm and EC = 8 cm

(ii) In \(\Delta\)ABC, DE || BC, AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)

\(\Rightarrow\) \(\frac{AD}{AB-AD}=\frac{AC-EC}{EC}\)

\(\Rightarrow\) \(\frac{AD}{13.5-AD}=\frac{11.9-5.1}{5.1}\)

\(\Rightarrow\) \(\frac{AD}{13.5-AD}=\frac{6.8}{5.1}\)

\(\Rightarrow\) 5.1AD = 91.8 – 6.8AD

\(\Rightarrow\) 11.9AD = 91.8

\(\Rightarrow\) AD = \(\frac{91.8}{11.9}\) = 7.7

Hence, AD = 7.7 cm

(iii) In \(\Delta\)ABC, DE || BC, AC = 6.6 cm, \(\frac{AD}{DB}=\frac{4}{7}\)

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)

\(\Rightarrow\) \(\frac{4}{7}=\frac{AE}{AC-AE}\)

\(\Rightarrow\) \(\frac{4}{7}=\frac{AE}{6.6-AE}\)

\(\Rightarrow\) 26.4 – 4AE = 7AE

\(\Rightarrow\) 26.4 = 11AE

\(\Rightarrow\) AE = \(\frac{26.4}{11}\) = 2.4 cm

Hence, AE = 2.4 cm

(iv) In \(\Delta\)ABC, DE || BC, Given \(\frac{AD}{AB}=\frac{8}{15}\), EC = 3.5 cm

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (By Thales theorem)

And AD = \(\frac{8}{15}\)AB [Given]

Therefore, \(\frac{AD}{AB-AD}=\frac{AE}{3.5}\) [AD = \(\frac{8}{15}\)AB]

\(\frac{\frac{8}{15}AB}{AB-\frac{8}{15}AB}=\frac{AE}{3.5}\)

\(\Rightarrow\) \(\frac{\frac{8}{15}AB}{\frac{7}{15}AB}=\frac{AE}{3.5}\)

\(\Rightarrow\) \(\frac{8}{7}=\frac{AE}{3.5}\)

\(\Rightarrow\) AE = \(\frac{3.5\times 8}{7}\)

\(\Rightarrow\) AE = 4 cm

Hence, AE = 4 cm

Question 2: D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC such that DE || BC.

Find the value of x, when:

(i) AD = x cm, DB = (x -2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

(ii) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

(iii) AD = (7x – 4) cm, AE = (5x – 2)cm, DB = (3x + 4) cm and EC = 3x cm.

Solution:

(i) D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC such that DE || BC, AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm, EC = (x – 1) cm

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (by thales theorem)

\(\Rightarrow\) \(\frac{x}{x-2}=\frac{x+2}{x-1}\)

\(\Rightarrow\) x(x – 1) = x^{2} – 4 \(\Rightarrow\) x^{2} – x = x^{2} – 4

\(\Rightarrow\) x = 4

(ii) In \(\Delta\)ABC, DE || BC, AD = 4 cm, DB = (x – 4) cm, AE = 8 cm, EC = (3x – 19) cm

\(\frac{AD}{AB}=\frac{AE}{EC}\) (by thales theorem)

\(\Rightarrow\) \(\frac{4}{x-4}=\frac{8}{3x-19}\)

\(\Rightarrow\) 4(3x – 19) = 8(x – 4)

\(\Rightarrow\) 12x – 76 = 8x – 32

\(\Rightarrow\) 4x = 44

\(\Rightarrow\) x = 11

\(\frac{AD}{DB}=\frac{AE}{EC}\) (by thales theorem)

\(\frac{7x-4}{3x+4}=\frac{5x-2}{3x}\)

Hence, x = 11

(iii) In \(\Delta\)ABC, DE || BC, AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm, EC = 3x cm

\(\Rightarrow\) 21x^{2} – 12x = 15x^{2} – 6x + 20x – 8

\(\Rightarrow\) 6x^{2} – 26x + 8 = 0

3x^{2} – 13x + 4 = 0

3x^{2} – 12x – x + 4 = 0

3x(x – 4) – 1(x – 4)

(x – 4)(3x – 1) = 0

x = 4 or x = \(\frac{1}{3}\)

If x = \(\frac{1}{3}\), length of sides become negative

Hence, x = 4

Question 3: D and E are points on the sides AB and AC respectively of a \(\Delta\)ABC. In each of the following cases, determine whether DE || BC or not.

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm.

(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm and AE = 4.2 cm.

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm and EC = 4 cm.

(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm.

Solution:

Given: A \(\Delta\)ABC, in which D and E are points on the sides AB and AC respectively.

To prove: DE || BC

Proof:

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

Since D and E are the points on AB and AC repectively.

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\)

\(\Rightarrow\) \(\frac{5.7}{9.5}=\frac{4.8}{8}\)

\(\Rightarrow\) 0.6 = 0.6

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (each equal to 0.6)

Hence, by the converse of Thales theorem DE || BC

(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm, AE = 4.2 cm

Since D and E are points on AB and AC respectively.

\(\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{AB-DB}{DB}=\frac{AE}{AC-AE}\)

\(\Rightarrow\) \(\frac{11.7-6.5}{6.5}=\frac{4.2}{11.2-4.2}\Rightarrow \frac{5.2}{6.5}\neq \frac{4.2}{7}\)

Hence, \(\frac{AD}{DB}\neq \frac{AE}{EC}\)

Hence, by the converse of Thales theorem DE is not parallel to BC.

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm, EC = 4 cm

Since D and E are the points on AB and AC respectively.

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) [by thales theorem]

\(\Rightarrow\) \(\frac{AD}{AB-AD}=\frac{AC-EC}{EC}\)

\(\Rightarrow\) \(\frac{6.3}{(10.8-6.3)}=\frac{(9.6-4.0)}{4}\)

\(\Rightarrow\) \(\frac{6.3}{4.5}=\frac{5.6}{4}\)

\(\Rightarrow\) 1.4 = 1.4

Therefore, \(\frac{AD}{DB}=\frac{AE}{EC}\) (each is equal to 1.4)

Hence by the converse of Thales theorem DE || BC

(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm, AC = 10 cm

Since D and E are points on the side AB and AC respectively.

\(\frac{AD}{DB}=\frac{AE}{EC}\) [by thales theorem]

\(\frac{7.2}{AB-AD}=\frac{6.4}{AC-AE}\)

\(\Rightarrow\) \(\frac{7.2}{12-7.2}=\frac{6.4}{10-6.4}\)

\(\Rightarrow\) \(\frac{7.2}{4.8}=\frac{3.4}{3.6}\)

but \(\frac{3}{2}\neq \frac{16}{9}\)

Hence, by the converse of Thales theorem DB is not parallel to BC

Question 4: In a \(\Delta\)ABC, AD is the bisector of \(\angle\)A.

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

(ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

Solution:

(i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm

Let BC = x

Now, DC = (BC – BD) = (x – 5.6) cm

In \(\Delta\)ABC, AD is the base of \(\angle\)A

So, by the angle bisector theorem, we have

\(\frac{BD}{DC}=\frac{AB}{AC}\)

\(\Rightarrow\) \(\frac{5.6}{x-5.6}=\frac{6.4}{8}\)

\(\Rightarrow\) 6.4x – 35.84 = 44.8

\(\Rightarrow\) 6.4x = 80.64

\(\Rightarrow\) x = 12.6

Hence, BC = 12.6 cm and DC = (12.6 – 5.6) cm = 7 cm

(ii) AB = 10 cm, AC = 14 cm, BC = 6 cm

Let BD = x,

DC = (BC – BD) = (6 – x) cm

In \(\Delta\)ABC, AD is the bisector of \(\angle\)A

So, By angle bisector theorem,

We have, \(\frac{BD}{DC}=\frac{AB}{AC}\)

\(\Rightarrow\) \(\frac{x}{6-x}=\frac{10}{14}\)

\(\Rightarrow\) 14x = 60 – 10x

\(\Rightarrow\) 24x = 60

\(\Rightarrow\) x = 2.5

Hence, BD = 2.5 cm and DC = (6 – 2.5) cm = 3.5 cm

(iii) AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm

DC = BC – BD = (6 – 3.2) cm = 2.8 cm

Let, AC = x,

In \(\Delta\)ABC, AD is the bisector of \(\angle\)A

So, by the angle bisector theorem we have

Therefore, \(\frac{BD}{DC}=\frac{AB}{AC}\)

\(\Rightarrow\) \(\frac{3.2}{2.8}=\frac{5.6}{x}\)

\(\Rightarrow\) x = \(\frac{5.6\times 2.8}{3.2}=4.9\) cm

Hence, AC = 4.9 cm

(iv) AB = 5.6 cm, AC = 4 cm, DC = 3 cm

Let BD = x,

In \(\Delta\)ABC, AD is the bisector of \(\angle\)A

So, by the angle bisector theorem we have

Therefore, \(\frac{BD}{DC}=\frac{AB}{AC}\)

\(\Rightarrow\) \(\frac{x}{3}=\frac{5.6}{4}\)

\(\Rightarrow\) x = \(\frac{5.6\times 3}{4}=4.2\) cm

Hence, BD = 4.2 cm

So BC = BD + AC = (4.2 + 3) cm

BC = 7.2 cm

Question 5: In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals interact at O. If AO = (5x – 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm, find the value of x.

Solution:

We know that CD || AB in trap ABCD and its diagonals interact at O.

Since the diagonals of a trapezium divide each other proportionally therefore, we have

\(\frac{AO}{OC}=\frac{BO}{OD}\) (by thales theorem)

\(\Rightarrow\) \(\frac{5x-7}{2x+1}=\frac{7x+1}{7x-5}\)

\(\Rightarrow\) (5x – 7)()7x – 5) = (7x + 1)(2x + 1)

\(\Rightarrow\) 35x^{2} – 25x – 49x + 35 = 14x^{2} + 7x + 2x + 1

\(\Rightarrow\) 35x^{2} – 14x^{2} – 25x – 49x – 7x – 2x + 35 – 1 = 0

\(\Rightarrow\) 21x^{2} – 83x + 34 = 0

Therefore, \(x=\frac{83\pm \sqrt{(83)^{2}-4\times 21\times 34}}{21\times 2}\)

= \(\frac{83\pm \sqrt{4033}}{42}\)

= \(\frac{83\pm 63.51}{42}\)

= \(\frac{146.51}{42}\) or \(\frac{19.49}{42}\) = 3.49 or 0.46

\(\Rightarrow\) x = 0.46 or 3.49,

But x \(\neq\) 0.46 as all sides should be positive.

Therefore, x = 3.49 cm

Question 6: \(\Delta\)ABC and \(\Delta\)DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD is drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.

Solution:

Given: \(\Delta\)ABC and \(\Delta\)DBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.

To Prove: QR || AD

Proof: In \(\Delta\)ABC

PQ || AB

\(\Rightarrow\) \(\frac{CP}{PB}=\frac{CQ}{QA}\) ——– (1)(by thales theorem)

In \(\Delta\)BCD, PR || BD

Therefore, \(\frac{CP}{PB}=\frac{CR}{RD}\) ——– (2)(by thales theorem)

From (1) & (2), we get:

\(\frac{CQ}{QA}=\frac{CR}{RD}\)

Hence, in \(\Delta\)ACD, Q and R the points in AC and CD such that

Therefore, \(\frac{CQ}{QA}=\frac{CR}{RD}\)

QR || AD (by the converse of Thales theorem)

Hence proved

Question 7: In the given figure, side BC of \(\Delta\)ABC is bisected at D and O is any point in AD. BO and CO produced meet AC and AB at E and 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 || BC.

Solution:

Given BD = CD and OD = DX

Join BX and CX

Thus, the diagonals of quad OBXC bisect each other

OBXC is a parallelogram

BX || CF and so, OF || BX

Similarly, CX || OE

In \(\Delta\)ABX, OF || BX

Therefore, \(\frac{AO}{AX}=\frac{AF}{AB}\) —– (1)

In \(\Delta\)ACX, OE || XC

Therefore, \(\frac{AO}{AX}=\frac{AE}{AC}\) —– (2)

From (1) & (2) we get:

\(\frac{AF}{AB}=\frac{AE}{AC}\)

Hence, EF || BC

Question 8: ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = \(\frac{1}{4}\)AC. If PQ produced meets BC at R, prove that R is the mid-point of BC.

Solution:

Given: ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac{1}{4}\)AC. PQ produced meets BC at R.

To prove: R is the mid-point of BC

Construction: Join BD

Proof: Since the diagonals of a parallelogram bisect each other at S such that

CS = ½ AC

Now, CS = ½ AC and CQ = \(\frac{1}{4}\)AC \(\Rightarrow\) CQ = ½ CS

Therefore, Q is the mid-point of CS

So, PQ || DS

Therefore, QR || SB

In \(\Delta\)CSB, Q is the mid-point of CS and QR || SB.

So R is the midpoint of BC.

Question 9: In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.

Solution:

Given: ABC is a triangle in which AB = AC. D and E are points on AB and AC respectively such that AD = AE

To prove: AB = AC (given)

\(\Rightarrow\) (AB – AD) = (AC – AE)

\(\Rightarrow\) DE = EC

\(\Rightarrow\) \(\frac{AD}{AE}=\frac{DB}{EC}\) (each equal to 1)

\(\Rightarrow\) DE || BC (by the converse of Thales’s theorem)

\(\Rightarrow\) \(\angle\)DEC + \(\angle\)ECB = 180^{o}

\(\Rightarrow\) \(\angle\)DEC + \(\angle\)CBD = 180^{o} [AB = AC \(\Rightarrow\) \(\angle\)C = \(\angle\)BB]

Quadrilateral BCEA is cyclic

Hence, the point B, C, E, D are concyclic