NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2#
Exercise 6.2 covers the Basic Proportionality Theorem (also known as Thales’ Theorem) and its converse. The BPT states that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
Basic Proportionality Theorem (BPT / Thales’ Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Conversely, if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
1. In figure, (i) and (ii), DE ∥ BC. Find EC in (i) and AD in (ii). #
(i) Find EC, given AD = 1.5 cm, DB = 3 cm, AE = 1 cm
DE ∥ BC, so by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\Rightarrow \frac{1.5}{3} = \frac{1}{EC}$$
$$\Rightarrow EC = \frac{1 \times 3}{1.5}$$
$$\Rightarrow EC = 2$$
∴ $EC = \mathbf{2}$ cm
(ii) Find AD, given DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm
DE ∥ BC, so by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\Rightarrow \frac{AD}{7.2} = \frac{1.8}{5.4}$$
$$\Rightarrow \frac{AD}{7.2} = \frac{1}{3}$$
$$\Rightarrow AD = \frac{7.2}{3}$$
$$\Rightarrow AD = 2.4$$
∴ $AD = \mathbf{2.4}$ cm
2. E and F are points on the sides PQ and PR respectively of a △PQR. For each of the following cases, state whether EF ∥ QR: #
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
$$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$$$$\Rightarrow 1.3 \neq 1.5$$
∴ $\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$, so EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
$$\frac{PF}{RF} = \frac{8}{9}$$$$\Rightarrow \frac{8}{9} = \frac{8}{9}$$
∴ $\dfrac{PE}{QE} = \dfrac{PF}{RF}$, so by the converse of BPT, EF ∥ QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
First we find EQ and FR:
$EQ = PQ - PE = 1.28 - 0.18 = 1.10$ cm
$FR = PR - PF = 2.56 - 0.36 = 2.20$ cm
$$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{9}{55}$$$$\Rightarrow \frac{9}{55} = \frac{9}{55}$$
∴ $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$, so by the converse of BPT, EF ∥ QR.
3. In the figure, if LM ∥ CB and LN ∥ CD, prove that $\dfrac{AM}{AB} = \dfrac{AN}{AD}$. #
Proof
In △ABC, LM ∥ CB, so by Basic Proportionality Theorem:
$$\frac{AM}{MB} = \frac{AL}{LC} \quad \cdots (1)$$In △ACD, LN ∥ CD, so by Basic Proportionality Theorem:
$$\frac{AN}{ND} = \frac{AL}{LC} \quad \cdots (2)$$From equation (1) and equation (2):
$$\frac{AM}{MB} = \frac{AN}{ND}$$Taking reciprocals:
$$\frac{MB}{AM} = \frac{ND}{AN}$$Adding 1 to both sides:
$$\frac{MB}{AM} + 1 = \frac{ND}{AN} + 1$$$$\Rightarrow \frac{MB + AM}{AM} = \frac{ND + AN}{AN}$$
$$\Rightarrow \frac{AB}{AM} = \frac{AD}{AN}$$
Taking reciprocals again:
$$\frac{AM}{AB} = \frac{AN}{AD}$$Hence proved. $\blacksquare$
4. In the figure, DE ∥ AC and DF ∥ AE. Prove that $\dfrac{BF}{FE} = \dfrac{BE}{EC}$. #
Proof
In △BAC, DE ∥ AC, so by Basic Proportionality Theorem:
$$\frac{BE}{EC} = \frac{BD}{DA} \quad \cdots (1)$$In △BAE, DF ∥ AE, so by Basic Proportionality Theorem:
$$\frac{BF}{FE} = \frac{BD}{DA} \quad \cdots (2)$$From equation (1) and equation (2):
$$\frac{BF}{FE} = \frac{BE}{EC}$$Hence proved. $\blacksquare$
5. In the figure, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR. #
Proof
In △POQ, DE ∥ OQ, so by Basic Proportionality Theorem:
$$\frac{PE}{EQ} = \frac{PD}{DO} \quad \cdots (1)$$In △POR, DF ∥ OR, so by Basic Proportionality Theorem:
$$\frac{PF}{FR} = \frac{PD}{DO} \quad \cdots (2)$$From equation (1) and equation (2):
$$\frac{PE}{EQ} = \frac{PF}{FR}$$∴ In △PQR, E and F divide PQ and PR in the same ratio, so by the converse of BPT, EF ∥ QR. $\blacksquare$
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. #
Proof
In △OPQ, AB ∥ PQ, so by Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OB}{BQ} \quad \cdots (1)$$In △OPR, AC ∥ PR, so by Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OC}{CR} \quad \cdots (2)$$From equation (1) and equation (2):
$$\frac{OB}{BQ} = \frac{OC}{CR}$$∴ In △OQR, B and C divide OQ and OR in the same ratio, so by the converse of BPT, BC ∥ QR. $\blacksquare$
7. Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). #
Proof
Let △ABC be a triangle in which D is the mid-point of AB and DE ∥ BC, meeting AC at E.
We need to prove that E is the mid-point of AC.
D is the mid-point of AB, so:
$$AD = DB$$$$\Rightarrow \frac{AD}{DB} = 1 \quad \cdots (1)$$
DE ∥ BC, so by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \cdots (2)$$From equation (1) and equation (2):
$$\frac{AE}{EC} = 1$$$$\Rightarrow AE = EC$$
∴ E is the mid-point of AC, so DE bisects the third side. $\blacksquare$
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). #
Proof
Let △ABC be a triangle in which D and E are the mid-points of AB and AC respectively.
We need to prove that DE ∥ BC.
D is the mid-point of AB, so:
$$AD = DB$$$$\Rightarrow \frac{AD}{DB} = 1 \quad \cdots (1)$$
E is the mid-point of AC, so:
$$AE = EC$$$$\Rightarrow \frac{AE}{EC} = 1 \quad \cdots (2)$$
From equation (1) and equation (2):
$$\frac{AD}{DB} = \frac{AE}{EC}$$∴ D and E divide AB and AC in the same ratio, so by the converse of BPT, DE ∥ BC. $\blacksquare$
9. ABCD is a trapezium in which AB ∥ DC and its diagonals intersect each other at point O. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$. #
Proof
Given: ABCD is a trapezium with AB ∥ DC. The diagonals AC and BD intersect at O.
Construction: Through O, draw EO ∥ DC, meeting AD at E. Then EO ∥ AB as well.
In △ADC, EO ∥ DC, so by Basic Proportionality Theorem:
$$\frac{AE}{ED} = \frac{AO}{OC} \quad \cdots (1)$$In △DAB, EO ∥ AB, so by Basic Proportionality Theorem:
$$\frac{DE}{EA} = \frac{DO}{OB}$$$$\Rightarrow \frac{AE}{ED} = \frac{BO}{OD} \quad \cdots (2)$$
From equation (1) and equation (2):
$$\frac{AO}{OC} = \frac{BO}{OD}$$$$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$$
Hence proved. $\blacksquare$
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$. Show that ABCD is a trapezium. #
Proof
Given: The diagonals AC and BD of quadrilateral ABCD intersect at O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.
We need to prove that AB ∥ DC.
Construction: Through O, draw EO ∥ AB, meeting AD at E.
From the given condition:
$$\frac{AO}{BO} = \frac{CO}{DO}$$$$\Rightarrow \frac{AO}{CO} = \frac{BO}{DO} \quad \cdots (1)$$
In △DAB, EO ∥ AB, so by Basic Proportionality Theorem:
$$\frac{DE}{EA} = \frac{DO}{OB}$$$$\Rightarrow \frac{AE}{ED} = \frac{BO}{DO} \quad \cdots (2)$$
From equation (1) and equation (2):
$$\frac{AE}{ED} = \frac{AO}{CO}$$So in △ADC, E and O divide AD and AC in the same ratio. By the converse of BPT:
$$EO \parallel DC$$We also have EO ∥ AB (by construction).
∴ AB ∥ DC, so ABCD is a trapezium. $\blacksquare$