NCERT Class 10 Maths – Exercise 6.2 Solutions

NCERT Class 10 Maths

Chapter 6 – Triangles | Exercise 6.2

(Rationalized Syllabus 2025-26)

💡 Key Theorem: BPT

Basic Proportionality Theorem (Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Q1

In Fig. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) Find EC
Given: In $\Delta ABC$, $DE || BC$.
By Basic Proportionality Theorem (BPT):
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Substitute values: $AD=1.5$, $DB=3$, $AE=1$.
$$\frac{1.5}{3} = \frac{1}{EC} \Rightarrow \frac{1}{2} = \frac{1}{EC} \Rightarrow EC = 2$$
✔️ EC = 2 cm
(ii) Find AD
Given: In $\Delta ABC$, $DE || BC$.
By BPT:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Substitute values: $DB=7.2$, $AE=1.8$, $EC=5.4$.
$$\frac{AD}{7.2} = \frac{1.8}{5.4} \Rightarrow \frac{AD}{7.2} = \frac{1}{3}$$
$$AD = \frac{7.2}{3} = 2.4$$
✔️ AD = 2.4 cm
Q2

E and F are points on the sides PQ and PR respectively of a $\Delta PQR$. 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
Check ratios:
$$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$$
$$\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{36}{24} = 1.5$$
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, EF is not parallel to QR.
✖️ EF is not parallel to QR
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Check ratios:
$$\frac{PE}{QE} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$$
$$\frac{PF}{RF} = \frac{8}{9}$$
Since $\frac{PE}{QE} = \frac{PF}{RF}$, by Converse of BPT, EF || QR.
✔️ EF || QR
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Here $EQ = PQ – PE = 1.28 – 0.18 = 1.10$.
And $FR = PR – PF = 2.56 – 0.36 = 2.20$.
Check ratios:
$$\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$$
$$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$$
Since ratios are equal, EF || QR.
✔️ EF || QR
Q3

In the figure, if LM || CB and LN || CD, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.

In $\Delta ABC$: Since $LM || CB$, by BPT:
$$\frac{AM}{AB} = \frac{AL}{AC} \quad \text{…(1)}$$
In $\Delta ADC$: Since $LN || CD$, by BPT:
$$\frac{AN}{AD} = \frac{AL}{AC} \quad \text{…(2)}$$
From (1) and (2), we get:
$$\frac{AM}{AB} = \frac{AN}{AD}$$
✔️ Hence Proved
Q4

In the figure, DE || AC and DF || AE. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.

In $\Delta ABC$: Since $DE || AC$, by BPT:
$$\frac{BD}{DA} = \frac{BE}{EC} \quad \text{…(1)}$$
In $\Delta ABE$: Since $DF || AE$, by BPT:
$$\frac{BD}{DA} = \frac{BF}{FE} \quad \text{…(2)}$$
From (1) and (2):
$$\frac{BF}{FE} = \frac{BE}{EC}$$
✔️ Hence Proved
Q5

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

In $\Delta POQ$: Since $DE || OQ$, by BPT:
$$\frac{PE}{EQ} = \frac{PD}{DO} \quad \text{…(1)}$$
In $\Delta POR$: Since $DF || OR$, by BPT:
$$\frac{PF}{FR} = \frac{PD}{DO} \quad \text{…(2)}$$
From (1) and (2):
$$\frac{PE}{EQ} = \frac{PF}{FR}$$
Now in $\Delta PQR$, since the sides are divided in the same ratio, by Converse of BPT:
✔️ EF || QR (Proved)
Q6

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.

In $\Delta OPQ$: Since $AB || PQ$, by BPT:
$$\frac{OA}{AP} = \frac{OB}{BQ} \quad \text{…(1)}$$
In $\Delta OPR$: Since $AC || PR$, by BPT:
$$\frac{OA}{AP} = \frac{OC}{CR} \quad \text{…(2)}$$
From (1) and (2):
$$\frac{OB}{BQ} = \frac{OC}{CR}$$
In $\Delta OQR$, since $\frac{OB}{BQ} = \frac{OC}{CR}$, by Converse of BPT:
✔️ BC || QR (Proved)
Q7

Using Theorem 6.1 (BPT), prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Consider $\Delta ABC$. Let $D$ be the midpoint of $AB$, so $AD = DB$.
Line $DE$ is drawn parallel to $BC$, intersecting $AC$ at $E$.
By BPT:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Since $AD = DB$, the ratio $\frac{AD}{DB} = 1$.
$$1 = \frac{AE}{EC} \Rightarrow AE = EC$$
Thus, $E$ is the midpoint of $AC$.
✔️ Hence Proved
Q8

Using Theorem 6.2 (Converse of BPT), prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Consider $\Delta ABC$. Let $D$ and $E$ be midpoints of $AB$ and $AC$.
So, $AD = DB \Rightarrow \frac{AD}{DB} = 1$.
And, $AE = EC \Rightarrow \frac{AE}{EC} = 1$.
Therefore:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
By Converse of BPT, since the line divides the two sides in the same ratio, it must be parallel to the third side.
✔️ DE || BC (Proved)
Q9

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.

Draw a line $OE || DC$ meeting $AD$ at $E$. (Since $DC || AB$, $OE || AB$ as well).
In $\Delta ADC$: Since $OE || DC$, by BPT:
$$\frac{AE}{ED} = \frac{AO}{CO} \quad \text{…(1)}$$
In $\Delta DAB$: Since $OE || AB$, by BPT:
$$\frac{AE}{ED} = \frac{BO}{DO} \quad \text{…(2)}$$
From (1) and (2):
$$\frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$$
✔️ Hence Proved
Q10

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that ABCD is a trapezium.

Given: $\frac{AO}{BO} = \frac{CO}{DO} \Rightarrow \frac{AO}{CO} = \frac{BO}{DO}$.
Draw a line $OE || AB$ meeting $AD$ at $E$.
In $\Delta DAB$: By BPT:
$$\frac{AE}{ED} = \frac{BO}{DO}$$
From given, substitute $\frac{BO}{DO}$ with $\frac{AO}{CO}$:
$$\frac{AE}{ED} = \frac{AO}{CO}$$
In $\Delta ADC$: Since $\frac{AE}{ED} = \frac{AO}{CO}$, by Converse of BPT, $OE || DC$.
Since $OE || AB$ (by construction) and $OE || DC$ (proved), then $AB || DC$.
Since one pair of opposite sides is parallel, ABCD is a trapezium.
✔️ Hence Proved
🎉 Exercise 6.2 Completed
learncbsehub.in