- Similar Figures
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- Similar Polygons
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- Basic Proportionally Theorem (or Thales Theorem)
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- Criteria for Similarity of Triangles
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- Different Criterion for similarity of the triangles
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- Areas of Similar Triangles
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- Pythagoras Theorem
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- Converse of Pythagoras Theorem

In this page we have *NCERT Solutions for Class 10 Mathematics:Chapter 6 Similar triangles* for
EXERCISE 6.2 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In the ΔABC , if DE || BC,

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

(i) In ΔABC, DE||BC

Now from Basic proportionality theorem

$\frac {AD}{DB} = \frac {AE}{EC}$

$ \frac {1.5}{3} = \frac {1}{EC}$

EC = 3/1.5

EC = 2 = 2 cm

(ii) In ΔABC, DE||BC

Now from Basic proportionality theorem

$\frac {AD}{DB} = \frac {AE}{EC}$

$ \frac {AD}{7.2} = \frac {1.8}{5.4}$

AD = 1.8×7.2/5.4 = 2.4 cm

E and F are points on the sides PQ and PR respectively of a ΔPQR. 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 = 0.63 cm

In ΔPQR, E and F are two points on side PQ and PR respectively.

(i) Given

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2,4 cm

Now the Ratio PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, $\frac {PE}{EQ} \neq \frac {PF}{FR}$

So, as per converse of Basic proportionality theorem

Hence, EF is not parallel to QR.

(ii) Given

PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm

Now the ratio PE/QE = 4/4.5 = 40/45 = 8/9

And PF/RF = 8/9

So, $\frac {PE}{EQ} = \frac {PF}{FR}$

So as per converse of Basic proportionality theorem

Hence, EF is parallel to QR.

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

Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm

And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55

And, PE/FR = 0.36/2.20 = 36/220 = 9/55

so, $\frac {PE}{EQ} = \frac {PF}{FR}$

So as per converse of Basic proportionality theorem

Hence, EF is parallel to QR.

In the below figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD

In the given figure, LM || CB

By using basic proportionality theorem, we get,

AM/MB = AL/LC ...

Similarly, LN || CD

By using basic proportionality theorem, we get,

AN/AD = AL/LC ...

From

AM/MB = AN/AD

In the below figure DE||AC and DF||AE. Prove that BF/FE = BE/EC

In ΔABC, DE || AC

By using Basic Proportionality Theorem

BD/DA = BE/EC ...

In ΔABC, DF || AE

By using Basic Proportionality Theorem

BD/DA = BF/FE ...

From equation

BE/EC = BF/FE

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

In ΔPQO, DE || OQ

By using Basic Proportionality Theorem

$ \frac {PD}{DO} = \frac {PE}{EQ}$ ...

In ΔPRO, DF||OR

By using Basic Proportionality Theorem

$ \frac {PD}{DO} = \frac {PF}{FR} $ ...

From equation

$ \frac {PE}{EQ} = \frac {PF}{FR}$

By converse of Basic Proportionality Theorem

So In ΔPQR, EF || QR.

In the below 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 ΔOPQ, AB || PQ

By using Basic Proportionality Theorem

$\frac {OA}{AP}= \frac {OB}{BQ}$ ...

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

$ \frac {OA}{AP}= \frac {OC}{CR}...

From equation

$ \frac {OB}{BQ} = \frac {OC}{CR}$

By converse of Basic Proportionality Theorem

So In ΔOQR, BC || QR.

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

ΔABC in which D is the mid-point of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

∴ AD=DB

⇒ AD/BD = 1 ...

In ΔABC, DE || BC,

By using Basic Proportionality Theorem

AD/DB = AE/EC

Now as AD/BD=1

AE/EC=1

So E is the mid-point of AC

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).

ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

D is the midpoint of AB (Given)

So AD=DB

Then AD/BD = 1 ...

Also, E is the mid-point of AC (Given)

So AE=EC

Then AE/EC = 1 [From equation

From equation

AD/BD = AE/EC

By converse of Basic Proportionality Theorem

Hence, DE || BC

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

Through O, draw EO || DC

Now as DC || AB, So, EO||AB

In ΔADC, we have

OE || DC (By Construction)

By using Basic Proportionality Theorem

$ \frac {AE}{ED} = \frac {AO}{CO}$ ...

In ΔABD, we have

OE || AB (Proved above)

By using Basic Proportionality Theorem

$ \frac {DE}{EA} = \frac {DO}{BO}$ ...

From equation

$\frac {AO}{CO} = \frac {BO}{DO}$

$ \frac {AO}{BO} = \frac {CO}{DO}$

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.

According to the question

Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that

$ \frac {AO}{BO} = \frac {CO}{DO}$ ---

Through O, draw line EO, where EO || AB, which meets AD at E.

In ΔDAB, we have

EO || AB

[By using Basic Proportionality Theorem

$ \frac {DE}{EA} = \frac {DO}{OB}$ ...

From

$ \frac {AO}{CO} = \fac {BO}{DO}$

Rearranging the equation,we get

$ \frac {CO}{AO} = \frac {BO}{DO}$

$ \frac {DO}{OB} = \frac {CO}{AO}$ ...

From equation

$ \frac {DE}{EA} = \frac {CO}{AO}$

By using converse of Basic Proportionality Theorem

Therefore, EO || DC

Now EO || AB

So AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.