# NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2

Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

Question 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 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 = 0.36 cm

Solution:
We have to show that EF || QR.
If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
(i) PE = 3.9cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Ratio are not same, so EF is not parallel to QR.

(ii) PE = 4cm, QE = 4.5cm, PF = 8cm and RF = 9cm

Therefore, the ratio are same, so EF || QR.

(iii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cm and PF = 0.36cm

Ratio are same, so EF || QR.

Question 3.
In the given figure, if LM || CB and LN || CD, Prove that $$\frac { AM }{ AB }$$ = $$\frac { AN }{ AD }$$

Solution:
Given a quadrilateral ABCD, in which LM || BC and LN || CD
In ∆ABC, LM || BC
$$\frac { AM }{ MB }$$ = $$\frac { AL }{ LC }$$ … (i)
[Basic proportionality theorem in fig.]
$$\frac { AN }{ AD }$$ = $$\frac { AL }{ LC }$$ … (2)
[Basis proportionality theorem]
From equations (1) and (2), we have

Question 4.
In the adjoining figure DE || AC and DF || AE. Prove that $$\frac { BF }{ FE }$$ = $$\frac { BE }{ EC }$$

Solution:
Given a quadrilateral ABC, in which DF || AE and DE || AC
In ∆ABE, we have
$$\frac { BD }{ AD }$$ = $$\frac { BE }{ FE }$$ … (i)
[Basic proportionality theorem in fig.]
In ∆ABC, we have
$$\frac { BD }{ AD }$$ = $$\frac { BE }{ EC }$$ … (2)
[Basis proportionality theorem]
From equations (1) and (2), we have
$$\frac { BE }{ FE }$$ = $$\frac { BE }{ EC }$$

Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.
Solution:
In the triangle PQD, ED || OQ.
So applying Thales theorem, we get,

Equation (iii) shows that EF intersects PQ and PR at E and F respectively in equal ratio. So, according to the converse of the basic proportionality theorem (or Thales theorem), EF || QR. Flence Proved.

Question 6.
In the given 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:
Given : O is any point inside ∆PQR. A is any point PO and B is any point on OQ and C is | any point on OR. AB || PQ and AC || PR.

To Prove: BC || QR
Proof: In ∆POQ, AB || PQ
∴ $$\frac { OA }{ AP }$$ = $$\frac { OB }{ BQ }$$ …(1) [Basic proportionality theorem]
In ∆POR, AC || PR
∴$$\frac { OA }{ AP }$$ = $$\frac { OC }{ OR }$$ …(2) [Basic proportionality theorem]
From equations (1) and (2)
$$\frac { OB }{ BQ }$$ = $$\frac { OC }{ OR }$$
But, we know that
In ∆OQR,if $$\frac { OB }{ BQ }$$ = $$\frac { OC }{ OR }$$
then we can say BC || QR.
[Converse of basic proportionality theorem]
Hence Proved.

Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE || BC
To Prove : E is the mid-point of AC.

Proof : We have to prove that E is the mid-point of AC. Suppose, let E be not the mid-point of AC. Let E’ be the mid-point of AC. Join DE’.
Now. in ∆ABC, D is the mid-point of AB and E’ is the mid-point of AC. Therefore, by Theorem (6.1).
DE || BC … (1)
Also DE || BC … (2)
From (1) and (2), we find that two intersecting lines DE and DE’ are both parallel to line BC. This is a contradiction to the parallel line axiom.
So, our assumption is wrong. Hence E is the mid-point of AC.

Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC
Construction : Produce the line segement DE to F, such that DE = EF. Join FC.

Proof: In ΔAED and CEF, we have
AE = CE
[∵ E is the mid-point of AC]
∠AED = ∠CEF
[Vertically opposite angle]
or DE = EF
So by SAS congruence
ΔAED ≅ ΔCEF
⇒ AD = CF … (1)
[corresponding of congruent triangle]
and ∠ADE = ∠CEF … (2)
Now, D is the mid-point of AB,
DB = CF … (3) [From (1)]
Now, DF intersects AD and FC at D and F such that
i.e., alternate interior angles are equal
DB || CF … (4)
From (3) and (4), DBCF is a quadrilateral such that one pair of sides are equal and parallel.
∴ DBCF is a parallelogram
⇒ DF || BC and DC = BC [opposite side of parallelogram]
But, D, E, F are collinear and DE = EF
∴ DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the O. Show that $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$

Solution:

Question 10.
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.
Solution:
Construction: Draw OF || AB, meeting AD in F.
To Prove: Quadrilateral ABCD is a trapezium.
Proof: In ABD, we have OF || AB

Hence Proved.

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