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
Since DE ∥ BC, by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\frac{1.5}{3} = \frac{1}{EC}$$$$EC = \frac{1 \times 3}{1.5} = \frac{3}{1.5} = 2 \text{ cm}$$∴ $EC = 2$ cm
(ii) Find AD, given DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm
Since DE ∥ BC, by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$$$\frac{AD}{7.2} = \frac{1.8}{5.4}$$$$\frac{AD}{7.2} = \frac{1}{3}$$$$AD = \frac{7.2}{3} = 2.4 \text{ cm}$$∴ $AD = 2.4$ 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
We need to check if $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$
$$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$$$$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$$Since $\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$, therefore EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
We need to check if $\dfrac{PE}{QE} = \dfrac{PF}{RF}$
$$\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$$$$\frac{PF}{RF} = \frac{8}{9}$$Since $\dfrac{PE}{QE} = \dfrac{PF}{RF}$, therefore EF ∥ QR (by converse of BPT).
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
We have: $EQ = PQ - PE = 1.28 - 0.18 = 1.10$ cm $FR = PR - PF = 2.56 - 0.36 = 2.20$ cm
$$\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$$$$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$$Since $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$, therefore EF ∥ QR (by converse of BPT).
3. In the figure, if LM ∥ CB and LN ∥ CD, prove that $\dfrac{AM}{AB} = \dfrac{AN}{AD}$. #
Proof
In △ABC, LM ∥ CB.
By Basic Proportionality Theorem:
$$\frac{AM}{MB} = \frac{AL}{LC} \quad \cdots (1)$$In △ACD, LN ∥ CD.
By Basic Proportionality Theorem:
$$\frac{AN}{ND} = \frac{AL}{LC} \quad \cdots (2)$$From (1) and (2):
$$\frac{AM}{MB} = \frac{AN}{ND}$$Adding 1 to both sides:
$$\frac{AM}{MB} + 1 = \frac{AN}{ND} + 1$$$$\frac{AM + MB}{MB} = \frac{AN + ND}{ND}$$$$\frac{AB}{MB} = \frac{AD}{ND}$$Taking reciprocals:
$$\frac{MB}{AB} = \frac{ND}{AD}$$Therefore:
$$1 - \frac{MB}{AB} = 1 - \frac{ND}{AD}$$$$\frac{AM}{AB} = \frac{AN}{AD}$$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.
By Basic Proportionality Theorem:
$$\frac{BE}{EC} = \frac{BD}{DA} \quad \cdots (1)$$In △BAE, DF ∥ AE.
By Basic Proportionality Theorem:
$$\frac{BF}{FE} = \frac{BD}{DA} \quad \cdots (2)$$From (1) and (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.
By Basic Proportionality Theorem:
$$\frac{PE}{EQ} = \frac{PD}{DO} \quad \cdots (1)$$In △POR, DF ∥ OR.
By Basic Proportionality Theorem:
$$\frac{PF}{FR} = \frac{PD}{DO} \quad \cdots (2)$$From (1) and (2):
$$\frac{PE}{EQ} = \frac{PF}{FR}$$Therefore, by the converse of Basic Proportionality Theorem applied to △PQR, 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.
By Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OB}{BQ} \quad \cdots (1)$$In △OPR, AC ∥ PR.
By Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OC}{CR} \quad \cdots (2)$$From (1) and (2):
$$\frac{OB}{BQ} = \frac{OC}{CR}$$Therefore, by the converse of Basic Proportionality Theorem applied to △OQR, 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.
Since D is the mid-point of AB:
$$AD = DB$$$$\Rightarrow \frac{AD}{DB} = 1 \quad \cdots (1)$$Since DE ∥ BC, by Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \cdots (2)$$From (1) and (2):
$$\frac{AE}{EC} = 1$$$$\Rightarrow AE = EC$$Therefore, E is the mid-point of AC. $\blacksquare$
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 mid-points of AB and AC respectively.
We need to prove that DE ∥ BC.
Since D is mid-point of AB: $AD = DB$
$$\Rightarrow \frac{AD}{DB} = \frac{1}{1} = 1 \quad \cdots (1)$$Since E is mid-point of AC: $AE = EC$
$$\Rightarrow \frac{AE}{EC} = \frac{1}{1} = 1 \quad \cdots (2)$$From (1) and (2):
$$\frac{AD}{DB} = \frac{AE}{EC}$$Therefore, by the converse of Basic Proportionality Theorem, 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.
Draw a line EF through O parallel to AB (and DC), meeting AD at E and BC at F.
In △DAB, EO ∥ AB (by construction).
By Basic Proportionality Theorem:
$$\frac{DE}{EA} = \frac{DO}{OB} \quad \cdots (1)$$In △ABC, OF ∥ AB (by construction).
By Basic Proportionality Theorem:
$$\frac{CF}{FB} = \frac{CO}{OA} \quad \cdots (2)$$Alternative direct proof:
Draw a line through A parallel to BD, meeting DC extended at E.
In △AEC, AB ∥ EC (since AB ∥ DC and E is on DC extended).
Wait — simpler approach:
In △AOB and △COD: $\angle AOB = \angle COD$ (vertically opposite angles) $\angle OAB = \angle OCD$ (alternate interior angles, AB ∥ DC) $\angle OBA = \angle ODC$ (alternate interior angles, AB ∥ DC)
So △AOB ~ △COD (AA similarity)
Therefore:
$$\frac{AO}{CO} = \frac{BO}{DO}$$$$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$$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: Diagonals AC and BD of quadrilateral ABCD intersect at O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.
We need to prove: ABCD is a trapezium (i.e., AB ∥ DC).
From the given condition:
$$\frac{AO}{BO} = \frac{CO}{DO}$$$$\Rightarrow \frac{AO}{CO} = \frac{BO}{DO} \quad \cdots (1)$$In △AOB and △COD:
$$\frac{AO}{CO} = \frac{BO}{DO}$$(from 1)
$\angle AOB = \angle COD$ (vertically opposite angles)
Therefore, by SAS similarity criterion:
$$\triangle AOB \sim \triangle COD$$$$\Rightarrow \angle OAB = \angle OCD$$These are alternate interior angles formed by AB and DC with transversal AC.
Since the alternate interior angles are equal, AB ∥ DC.
Therefore, ABCD is a trapezium. $\blacksquare$