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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 180°

**Axiom 6.2 :**If the sum of two adjacent angles is 180°, 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 = 180° … (1)

∠AOD + ∠BOD = 180° … (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 =180°(ii) ∠3 + ∠ 6 =180°

**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 180º.**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 ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find∠ BOE and reflex ∠ COE.**

Answer:

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

∠AOC + ∠COE + ∠BOE = 180°

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

= > 70° + ∠COE = 180°

= > ∠ COE = 180° -70°

= > ∠ COE = 110°

Also

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

∠COE + ∠BOE + ∠BOD = 180°

Put ∠COE = 110° and ∠ BOD = 40° we get

110° + ∠BOE + 40° = 180°

= > ∠BOE = 180° - 110° - 40°

**= > ∠BOE = 30° **

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

**Answer:**

Let ∠ a = 2x, then ∠b = 3x

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

∠XOM + ∠MOP + ∠POY = 180**°**

∠b + ∠a + ∠POY = 180**°**

given that ∠POY = 90**°**

plug this value we get

3x + 2x + 90**°** = 180**°**

5x = 90**°**

x = 18**°**

a = 2x = 2 × 18 = 36**°**

b = 3x = 3 ×18 = 54**°**

MN is a straight line.

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

so that ∠b + ∠c = 180**°**

54**°** + ∠c = 180**°**

∠c = 180**°** − 54**°** = 126**°**

∠ c = 126**°**

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

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

∠PQS + ∠PQR = 180° … (1)

And

∠PRT + ∠PRQ = 180° … (2)

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

∠PQS + ∠PQR = ∠PRT + ∠PRQ … (3)

But given that ∠PQR = ∠PRQ

Plug the value we get

∠PQS + ∠PRQ =∠PRT + ∠PRQ

∠PQS = ∠PRT + ∠PRQ - ∠PRQ

∠PQS = ∠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 360°

Hence

∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°

=> x + y + w + z = 360°

=> x + y + x + y = 360°

Given that x + y = w + z

Plug the value we get

=> 2w + 2z = 360°

=> 2(w + z) = 360°

w + z = 180° (linear pair)

or ∠BOD + ∠AOD = 180°

If the sum of two adjacent angles is 180°, 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 ∠ ROS = 1/2(∠ QOS – ∠ POS).**

**Solution **

It is given that OR is perpendicular to PQ

So that ∠POR = 90°

sum of angle in linear pair always equal to 180°

∠POS + ∠SOR + ∠POR = 180°

Plug ∠POR = 90°

90°+∠SOR + ∠POR = 180°

∠SOR + ∠POR = 90°

∠ROS = 90° − ∠POS … (1)

∠QOR = 90°

Given that OS is another ray lying between rays

OP and OR so that

∠QOS − ∠ROS = 90°

∠ROS = ∠QOS − 90° … (2)

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

2 ∠ROS = ∠QOS − ∠POS

∠ROS = 1/2(∠QOS − ∠POS)** **

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

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

Hence,

∠QYP = ∠ZYQ

PX is straight line

sum of angle in linear pair always equal to 180°

∠XYZ + ∠ZYQ + ∠QYP = 180°

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

And ∠QYP = ∠ZYQ

∠ 64° + 2∠QYP = 180°

∠2∠QYP = 180° − 64° = 116°

Divide by 2 we get

∠QYP = 58°

Also, ∠ZYQ = ∠QYP = 58°

Using angle of reflection

∠QYP = 360° − 58° = 302°

∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58°

= 122°

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

**Solution:**

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

x = 180° -50°

x= 130°

y is vertical opposite angle of 130° so that

y = 130°

**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 ∠ y = 3a

Then ∠z = 7a

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

∠x = ∠z …(1)

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

So that

x + y = 180°

plug the value of x from equation (1)

z + y = 180°

plug the value of z and y we get

7a + 3a = 180°

10 a = 180°

a = 180°/10

a = 18

y = 3a = 3x18 = 54°

z = 7a = 7x18 = 126°

x = z = 126°

**Question 3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.**

**Solution :**

∠GEF + ∠FEG = ∠GED

EF ⊥ CD so that ∠ FED = 90°

And given that ∠ GED = 126°

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

∠GEF + 90° = 126°

∠GEF = 126°- 90°

∠GEF = 36°

AB || CD, EF ⊥ CD so ∠EFG = 90°

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

∠GEF + ∠EFG + ∠FGE = 180°

Plug the values we get

36° + 90° + ∠FGE = 180°

∠FGE = 180° – 36° - 90°

∠FGE = 54°

AB is a straight line so that use linear pair property

∠AGE +∠FGE = 180°

Plug the value of ∠FGE = 54°

We get

∠AGE +54° = 180°

∠AGE = 180° - 54°

∠AGE = 126°

**Question 4. In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ 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 = 180°

So that

∠ PQR + ∠ QRX = 180°

Given that ∠ PQR= 110°

110° + ∠QRX = 180°

∠QRX = 180° -110°

∠QRX = 70°

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

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

Also

130° + ∠SRY = 180°

∠SRY = 50°

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

∠QRX + ∠QRS + ∠SRY = 180°

70° + ∠QRS + 50° = 180°

∠QRS = 180° − 120°

= 60°

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

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

∠APR = ∠PRD

50° + y = 127°

y = 127° − 50°

y = 77°

use same property of Alternate interior angles

∠APQ = ∠PQR

50° = x

∠ x = 50° and y = 77°

**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 ⊥ PQ and CN ⊥ RS.

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

Use the property of Alternate interior angles

∠2 = ∠3 … (1)

∠ABC = ∠1 + ∠2

But ∠1 = ∠2 so that

∠ABC = ∠2 + ∠2

∠ABC = 2∠2

Similarly

∠BCD = ∠3 + ∠4

But ∠3 = ∠4 so that

∠ BCD = ∠3 + ∠3

∠ BCD = 2∠3

From equation first

∠ABC = ∠DCB

These are alternate angles so that AB || CD

Hence proved

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