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Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.2
  • Last Updated : 03 Mar, 2021

Theorem 6.1 :

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.

Theorem 6.2 :

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

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



Solution:

(i) Here, In △ ABC,

DE || BC

So, according to Theorem 6.1

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

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

⇒EC =\frac{3×1}{1.5}

EC = 2 cm



Hence, EC = 2 cm.

(ii) Here, In △ ABC,

So, according to Theorem 6.1 , if DE || BC

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

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

⇒AD =\frac{1.8×7.2}{5.4}

AD = 2.4 cm

Hence, AD = 2.4 cm.

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

Solution:

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{3.9}{3} = 1.3 ………………………(i)

\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5 ………………………(ii)

As,\frac{PE}{EQ} ≠ \frac{PF}{FR}

Hence, EF is not parallel to QR.

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

Solution:

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{4}{4.5} = \frac{8}{9} ………………………(i)

\frac{PF}{FR} = \frac{8}{9} ………………………(ii)

As,\frac{PE}{EQ} = \frac{PF}{FR}

Hence, EF is parallel to QR.

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

Solution:

EQ = PQ – PE = 1.28 – 0.18 = 1.1

and, FR = PR – PF = 2.56 – 0.36 = 2.2

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{0.18}{1.1} = \frac{9}{55} ………………………(i)

\frac{PF}{FR} = \frac{0.36}{2.2} = \frac{9}{55} ………………………(ii)

As,\frac{PE}{EQ} = \frac{PF}{FR}

Hence, EF is parallel to QR.

Question 3. In Figure, if LM || CB and LN || CD, prove that

\frac{AM}{AB} = \frac{AN}{AD}

Solution:

Here, In △ ABC,

According to Theorem 6.1, if LM || CB

then,\frac{AM}{AB} = \frac{AL}{AC} …………………………….(I)

and, In △ ADC,

According to Theorem 6.1, if LN || CD

then,\frac{AN}{AD} = \frac{AL}{AC} …………………………….(II)

From (I) and (II), we conclude that

\frac{AM}{AB} = \frac{AN}{AD}

Hence Proved !!

Question 4. In Figure, DE || AC and DF || AE. Prove that

\frac{BF}{FE} = \frac{BE}{EC}

Solution:

Here, In △ ABC,

According to Theorem 6.1, if DE || AC

then,\frac{BD}{DA} = \frac{BE}{EC} …………………………….(I)

and, In △ ABE,

According to Theorem 6.1, if DF || AE

then,\frac{BD}{DA} = \frac{BF}{FE} …………………………….(II)

From (I) and (II), we conclude that

\frac{BF}{FE} = \frac{BE}{EC}

Hence Proved !!

Question 5. In Figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

Here, In △ POQ,

According to Theorem 6.1, if DE || OQ

then,\frac{PE}{EQ} = \frac{PD}{DO} …………………………….(I)

and, In △ POR,

According to Theorem 6.1, if DF || OR

then,\frac{PD}{DO} = \frac{PF}{FR} …………………………….(II)

From (I) and (II), we conclude that

\frac{PE}{EQ} = \frac{PF}{FR} ………………………………(III)

According to Therem 6.2 and eqn. (III)

EF || QR, in △ PQR

Hence Proved !!

Question 6. In 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:

Here, In △ POQ,

According to Theorem 6.1, if AB || PQ

then,\frac{OA}{AP} = \frac{OB}{BQ} …………………………….(I)

and, In △ POR,

According to Theorem 6.1, if AC || PR

then,\frac{OA}{AP} = \frac{OC}{CR} …………………………….(II)

From (I) and (II), we conclude that

\frac{OB}{BQ} = \frac{OC}{CR} ………………………………(III)

According to Therem 6.2 and eqn. (III)

BC || QR, in △ OQR

Hence Proved !!

Question 7. Using Theorem 6.1, 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, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E

So, DE || BC.

We have to prove that E is the mid point of AC.

As, AD=DB

\frac{AD}{DB} = 1 …………………………. (I)

Here, In △ ABC,

According to Theorem 6.1, if DE || BC

then,\frac{AD}{DB} = \frac{AE}{EC} …………………………….(II)

From (I) and (II), we conclude that

\frac{AD}{DB} = \frac{AE}{EC} = 1

\frac{AE}{EC} = 1

AE = EC

E is the midpoint of AC.

Hence proved !!

Question 8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution:

Given, in ΔABC, D and E are the mid points of AB and AC respectively

AD=BD and AE=EC.

We have to prove that: DE || BC.

As, AD=DB

\frac{AD}{DB} = 1 …………………………. (I)

and, AE=EC

\frac{AE}{EC} = 1 …………………………. (II)

From (I) and (II), we conclude that

\frac{AD}{DB} = \frac{AE}{EC} = 1 ……………….(III)

According to Therem 6.2 and eqn. (III)

DE || BC, in △ ABC

Hence Proved !!

Question 9. 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}

Solution:

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

Here, In △ ADB,

According to Theorem 6.1, if AB || EO

then,\frac{AE}{ED} = \frac{OB}{OD} …………………………….(I)

and, In △ ADC,

According to Theorem 6.1, if AC || PR

then,\frac{AE}{ED} = \frac{AO}{OC} …………………………….(II)

From (I) and (II), we conclude that

\frac{AO}{OC} = \frac{OB}{OD}

After rearranging, we get

\frac{AO}{OB} = \frac{OC}{OD}

Hence Proved !!

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:

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

Here, In △ ADB,

According to Theorem 6.1, if AB || EO

then,\frac{AE}{ED} = \frac{OB}{OD} …………………………….(I)

\frac{AO}{OB} = \frac{OC}{OD} (Given)

\frac{AO}{OC} = \frac{OB}{OD} (After rearranging) ………………………………..(II)

From (I) and (II), we conclude that

\frac{AE}{ED} = \frac{AO}{OC} ………………………………..(III)

According to Therem 6.2 and eqn. (III)

EO || DC and also EO || AB

⇒ AB || DC.

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

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