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**Axioms of Chapter 6**

**Axiom 6.1 :**If a ray stands on a line, then the sum of two adjacent angles so

formed is 180Â°

**Axiom 6.2 :**If the sum of two adjacent angles is 180Â°, then the non-common arms

of the angles form a line

**Axiom 6.3 :**If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

Axiom 6.3 is also referred to as the **corresponding angles axiom**.

**Axiom 6.4 ****: **If a transversal intersects two lines such that a pair of corresponding

angles is equal, then the two lines are parallel to each other.

**Theorem 6. 1 :**If two lines intersect each other, then the vertically opposite angles are equal **Given :-**two lines intersect each other**To prove :-** ∠AOC = ∠BOD and ∠AOD = ∠BOC

**Proof :- AB and CD be two lines intersecting at O as shown in Fig 6 8 They lead to two pairs ofvertically opposite angles, namely, (i) ∠AOC and ∠ BOD (ii) ∠ AOD and ∠ BOC **

We need to prove that ∠AOC = ∠BOD

and ∠AOD = ∠BOC

Now, ray OA stands on line CD

From (Linear pair axiom)

∠AOC + ∠AOD = 180Â° â€¦ (1)

∠AOD + ∠BOD = 180Â° â€¦ (2)

From (1) and (2), we can write

∠AOC + ∠AOD = ∠AOD + ∠BOD

This implies that ∠AOC = ∠BOD

Similarly, it can be proved that ∠AOD = ∠BOC

**Parallel Lines and a Transversal**

∠1, ∠ 2, ∠7 and ∠ 8 are called **exterior angles**,

∠ 3, ∠ 4, ∠ 5 and ∠6 are called **interior angles**

(a) **Corresponding angles :**

(i) ∠1 = ∠ 5 (ii) ∠ 2 = ∠ 6 (iii) ∠ 4 = ∠ 8 (iv) ∠ 3 = ∠ 7

(b) **Alternate interior angles**** :**

(i) ∠ 4 = ∠ 6 (ii) ∠ 3 = ∠ 5

(c) **Alternate exterior angles:**

(i) ∠1 = ∠ 7 (ii) ∠ 2 = ∠8

(d) **Interior angles on the same side of the transversal****:**

(i) ∠4 + ∠5 =180Â°(ii) ∠3 + ∠ 6 =180Â°

**Theorem 6.2 :**If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.**Theorem 6.4 ****: **If a transversal intersects two parallel lines, then each pair ofinterior angles on the same side of the transversal is supplementary.

**Theorem 6.5 :**If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.**Theorem 6.6 :**Lines which are parallel to the same line are parallel to each other.**Theorem 6.7**** : **The sum of the angles of a triangle is 180Âº.**Theorem 6.8 :**If a side of a triangle is produced, then the exterior angle soformed is equal to the sum of the two interior opposite angles.

**Question 1. In Fig. 6.13, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, findâˆ BOE and reflex âˆ COE.**

Answer:

(sum of angle in linear pair always equal to 180 )

âˆ AOC + âˆ COE + âˆ BOE = 180Â°

Given that âˆ AOC + âˆ BOE = 70Â° plug this value we get

= > 70Â° + âˆ COE = 180Â°

= > âˆ COE = 180Â° -70Â°

= > âˆ COE = 110Â°

Also

(sum of angle in linear pair always equal to 180Â° )

âˆ COE + âˆ BOE + âˆ BOD = 180Â°

Put âˆ COE = 110Â° and âˆ BOD = 40Â° we get

110Â° + âˆ BOE + 40Â° = 180Â°

= > âˆ BOE = 180Â° - 110Â° - 40Â°

**= > âˆ BOE = 30Â° **

**Question 2. In Fig. 6.14, lines XY and MN intersect at O. If âˆ POY = 90Â° and a : b = 2 : 3, find c.**

**Answer:**

Let âˆ a = 2x, then âˆ b = 3x

(sum of angle in linear pair always equal to 180**Â°** )

âˆ XOM + âˆ MOP + âˆ POY = 180**Â°**

âˆ b + âˆ a + âˆ POY = 180**Â°**

given that âˆ POY = 90**Â°**

plug this value we get

3x + 2x + 90**Â°** = 180**Â°**

5x = 90**Â°**

x = 18**Â°**

a = 2x = 2 Ã— 18 = 36**Â°**

b = 3x = 3 Ã—18 = 54**Â°**

MN is a straight line.

sum of angle in linear pair always equal to 180**Â°**

so that âˆ b + âˆ c = 180**Â°**

54**Â°** + âˆ c = 180**Â°**

âˆ c = 180**Â°** âˆ’ 54**Â°** = 126**Â°**

âˆ c = 126**Â°**

**Question 3. In Fig. 6.15, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT.**

**Solution: **ST is a straight line and sum of angle in linear pair always equal to 180

âˆ PQS + âˆ PQR = 180Â° â€¦ (1)

And

âˆ PRT + âˆ PRQ = 180Â° â€¦ (2)

From equation (1) and (2).we get:

âˆ PQS + âˆ PQR = âˆ PRT + âˆ PRQ â€¦ (3)

But given that âˆ PQR = âˆ PRQ

Plug the value we get

âˆ PQS + âˆ PRQ =âˆ PRT + âˆ PRQ

âˆ PQS = âˆ PRT + âˆ PRQ - âˆ PRQ

âˆ PQS = âˆ PRT

Hence proved

**Question 4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.**

**Solution:** Sum of all angles in a circle always 360Â°

Hence

âˆ AOC + âˆ BOC + âˆ DOB + âˆ AOD = 360Â°

=> x + y + w + z = 360Â°

=> x + y + x + y = 360Â°

Given that x + y = w + z

Plug the value we get

=> 2w + 2z = 360Â°

=> 2(w + z) = 360Â°

w + z = 180Â° (linear pair)

or âˆ BOD + âˆ AOD = 180Â°

If the sum of two adjacent angles is 180Â°, then the non-common arms of the angles form a line

Hence AOB is a line.

**Question 5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between raysOP and OR. Prove that âˆ ROS = 1/2(âˆ QOS â€“ âˆ POS).**

**Solution **

It is given that OR is perpendicular to PQ

So that âˆ POR = 90Â°

sum of angle in linear pair always equal to 180Â°

âˆ POS + âˆ SOR + âˆ POR = 180Â°

Plug âˆ POR = 90Â°

90Â°+âˆ SOR + âˆ POR = 180Â°

âˆ SOR + âˆ POR = 90Â°

âˆ ROS = 90Â° âˆ’ âˆ POS â€¦ (1)

âˆ QOR = 90Â°

Given that OS is another ray lying between rays

OP and OR so that

âˆ QOS âˆ’ âˆ ROS = 90Â°

âˆ ROS = âˆ QOS âˆ’ 90Â° â€¦ (2)

On adding equations (1) and (2), we obtain

2 âˆ ROS = âˆ QOS âˆ’ âˆ POS

âˆ ROS = 1/2(âˆ QOS âˆ’ âˆ POS)** **

**Question 6. It is given that âˆ XYZ = 64Â° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects âˆ ZYP, find âˆ XYQ and reflex âˆ QYP.**

**Solution: **It is given that line YQ bisects âˆ PYZ.

Hence,

âˆ QYP = âˆ ZYQ

PX is straight line

sum of angle in linear pair always equal to 180Â°

âˆ XYZ + âˆ ZYQ + âˆ QYP = 180Â°

Give that so plug the value we get âˆ XYZ = 64Â°

And âˆ QYP = âˆ ZYQ

âˆ 64Â° + 2âˆ QYP = 180Â°

âˆ 2âˆ QYP = 180Â° âˆ’ 64Â° = 116Â°

Divide by 2 we get

âˆ QYP = 58Â°

Also, âˆ ZYQ = âˆ QYP = 58Â°

Using angle of reflection

âˆ QYP = 360Â° âˆ’ 58Â° = 302Â°

âˆ XYQ = âˆ XYZ + âˆ ZYQ

= 64Â° + 58Â°

= 122Â°

**Question 1. In Fig. 6.28, find the values of x and y and then show that AB || CD. **

**Solution:**

50Â° + x = 180Â° (sum of angle in linear pair is 180Â° )

x = 180Â° -50Â°

x= 130Â°

y is vertical opposite angle of 130Â° so that

y = 130Â°

**Question 2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Solution:**

Given that y : z = 3 : 7

Let âˆ y = 3a

Then âˆ z = 7a

âˆ x and âˆ z are alternate interior angles of parallel lines so that

âˆ x = âˆ z â€¦(1)

Sum of interior angle on the same side of the transversal is always = 180Â°

So that

x + y = 180Â°

plug the value of x from equation (1)

z + y = 180Â°

plug the value of z and y we get

7a + 3a = 180Â°

10 a = 180Â°

a = 180Â°/10

a = 18

y = 3a = 3x18 = 54Â°

z = 7a = 7x18 = 126Â°

x = z = 126Â°

**Question 3. In Fig. 6.30, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE.**

**Solution :**

âˆ GEF + âˆ FEG = âˆ GED

EF âŠ¥ CD so that âˆ FED = 90Â°

And given that âˆ GED = 126Â°

**Plug the values we get**** **

âˆ GEF + 90Â° = 126Â°

âˆ GEF = 126Â°- 90Â°

âˆ GEF = 36Â°

AB || CD, EF âŠ¥ CD so âˆ EFG = 90Â°

Use angle sum property of triangle (sum is 180Â°)

âˆ GEF + âˆ EFG + âˆ FGE = 180Â°

Plug the values we get

36Â° + 90Â° + âˆ FGE = 180Â°

âˆ FGE = 180Â° â€“ 36Â° - 90Â°

âˆ FGE = 54Â°

AB is a straight line so that use linear pair property

âˆ AGE +âˆ FGE = 180Â°

Plug the value of âˆ FGE = 54Â°

We get

âˆ AGE +54Â° = 180Â°

âˆ AGE = 180Â° - 54Â°

âˆ AGE = 126Â°

**Question 4. In Fig. 6.31, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS. [Hint : Draw a line parallel to ST throughpoint R.] **

**Solution:** Let us draw a parallel line XY to PQ || ST and passing through point R.

Sum of interior angle on the same side of the transversal is always = 180Â°

So that

âˆ PQR + âˆ QRX = 180Â°

Given that âˆ PQR= 110Â°

110Â° + âˆ QRX = 180Â°

âˆ QRX = 180Â° -110Â°

âˆ QRX = 70Â°

Sum of interior angle on the same side of the transversal is always = 180Â°

âˆ RST + âˆ SRY = 180Â° (Co-interior angles on the same side of transversal SR)

Also

130Â° + âˆ SRY = 180Â°

âˆ SRY = 50Â°

XY is a straight line. Use property of linear pair we get

âˆ QRX + âˆ QRS + âˆ SRY = 180Â°

70Â° + âˆ QRS + 50Â° = 180Â°

âˆ QRS = 180Â° âˆ’ 120Â°

= 60Â°

**Question 5. In Fig. 6.32, if AB || CD, âˆ APQ = 50Â° and âˆ PRD = 127Â°, find x and y.**

**Solution: **Use property of Alternate interior angles

âˆ APR = âˆ PRD

50Â° + y = 127Â°

y = 127Â° âˆ’ 50Â°

y = 77Â°

use same property of Alternate interior angles

âˆ APQ = âˆ PQR

50Â° = x

âˆ x = 50Â° and y = 77Â°

**Question 6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove thatAB || CD.**

**Solution:** Let draw BM âŠ¥ PQ and CN âŠ¥ RS.

Given that PQ || RS so that BM || CN

Use the property of Alternate interior angles

âˆ 2 = âˆ 3 â€¦ (1)

âˆ ABC = âˆ 1 + âˆ 2

But âˆ 1 = âˆ 2 so that

âˆ ABC = âˆ 2 + âˆ 2

âˆ ABC = 2âˆ 2

Similarly

âˆ BCD = âˆ 3 + âˆ 4

But âˆ 3 = âˆ 4 so that

âˆ BCD = âˆ 3 + âˆ 3

âˆ BCD = 2âˆ 3

From equation first

âˆ ABC = âˆ DCB

These are alternate angles so that AB || CD

Hence proved

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