**Axiom 3:** If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

Here, Exterior angles are ∠1, ∠2, ∠7 and ∠8

Interior angles are ∠3, ∠4, ∠5 and ∠6

Corresponding angles are ∠

(i) ∠1 and ∠5

(ii) ∠2 and ∠6

(iii) ∠4 and ∠8

(iv) ∠3 and ∠7

**Axiom 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.

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

Alternate Interior Angles: (i) ∠4 and ∠6 and (ii) ∠3 and ∠5

Alternate Exterior Angles: (i) ∠1 and ∠7 and (ii) ∠2 and ∠8

If a transversal intersects two parallel lines then each pair of alternate interior and exterior angles are equal.

Alternate Interior Angles: (i) ∠4 = ∠6 and (ii) ∠3 = ∠5

Alternate Exterior Angles: (i) ∠1 = ∠7 and (ii) ∠2 = ∠8

Interior angles on the same side of the transversal line are called the consecutive interior angles or allied angles or co-interior angles. They are as follows: (i) ∠4 and ∠5, and (ii) ∠3 and ∠6

**Theorem 2** If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

**Solution:** Given: Let PQ and RS are two parallel lines and AB be the transversal which intersects them on L and M respectively.

To Prove: ∠PLM = ∠SML

And ∠LMR = ∠MLQ

**Proof:** ∠PLM = ∠RMB ………….equation (i) (Corresponding ngles)

∠RMB = ∠SML ………….equation (ii) (vertically opposite angles)

From equation (i) and (ii)

∠PLM = ∠SML

Similarly, ∠LMR = ∠ALP ……….equation (iii) (corresponding angles)

∠ALP = ∠MLQ …………equation (iv) (vertically opposite angles)

From equation (iii) and (iv)

∠LMR = ∠MLQ Proved

**Theorem 3:** If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

**Solution:** Given: - A transversal AB intersects two lines PQ and RS such that

∠PLM = ∠SML

To Prove: PQ ||RS

Use same figure as in Theorem 2.

Proof: ∠PLM = ∠SML ……………equation (i) (Given)

∠SML = ∠RMB …………equation (ii) (vertically opposite angles)

From equations (i) and (ii);

∠PLM = ∠RMB

But these are corresponding angles.

We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines ate parallel to each other.

Hence, PQ║RS Proved.

**Theorem 4:** If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

**Solution:**Solution:

**Given:** Transversal EF intersects two parallel lines AB and CD at G and H respectively.

To Prove: ∠1 + ∠4 = 180° and ∠2 + ∠3 = 180°

Proof: ∠2 + ∠5 = 180° ………equation (i) (Linear pair of angles)

But ∠5 = ∠3 ……………equation (ii) (corresponding angles)

From equations (i) and (ii),

∠2 + ∠3 = 180°

Also, ∠3 + ∠4 = 180° ………equation (iii) (Linear pair)

But ∠3 = ∠1 …………..equation (iv) (Alternate interior angles)

From equations (iii) and (iv)

∠1 + ∠4 = 180° and ∠2 + ∠3 = 180° Proved

**Theorem 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.

**Solution:**

**Given:** A transversal EF intersects two lines AB and CD at P and Q respectively.

To Prove: AB ||CD

Proof: ∠1 + ∠2 = 180° ………..equation (i) (Given)

∠1 + ∠3 = 180° …………..equation (ii) (Linear Pair)

From equations (i) and (ii)

∠1 + ∠2 = ∠1 + ∠3

Or, ∠1 + ∠2 - ∠1 = ∠3

Or, ∠2 = ∠3

But these are alternate interior angles. We know that if a transversal intersects two lines such that the pair of alternate interior angles are equal, then the lines are parallel.

Hence, AB║CD Proved.

**Theorem 6:** Lines which are parallel to the same line are parallel to each other.

**Solution:**

**Given:** Three lines AB, CD and EF are such that AB║CD, CD║EF.**To Prove:** AB║EF.**Construction:** Let us draw a transversal GH which intersects the lines AB, CD and EF at P, Q and R respectively.**Proof:** Since, AB║CD and GH is the transversal. Therefore,

∠1 = ∠2 ………….equation (i) (corresponding angles)

Similarly, CD ||EF and GH is transversal. Therefore;

∠2 = ∠3 ……………equation (ii) (corresponding angles)

From equations (i) and (ii)

∠1 = ∠3

But these are corresponding angles.

We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines ate parallel to each other.

Hence, AB║ EF Proved.

**Theorem 7:** The sum of the angles of a triangle is 180º.

**Solution:**

**Given:** Δ ABC.

To Prove: ∠1 + ∠2 + ∠3 = 180°

Construction: Let us draw a line m though A, parallel to BC.

Proof: BC ||m and AB and AC are its transversal.

Hence, ∠1 = ∠4 …………….equation (i) (alternate interior angles)

∠2 = ∠5 ………..equation (ii) (alternate interior angles)

By adding equation (i) and (ii)

∠1 + ∠2 = ∠4 + ∠5 ………..equation (iii)

Now, by adding ∠3 to both sides of equation (iii), we get

∠1 + ∠2 + ∠3 = ∠4 + ∠5 + ∠3

Since, ∠4 + ∠5 + ∠ = 180° (Linear group of angle)

Hence, ∠1 + ∠2 + ∠3 = 180°

Hence Proved.

**Theorem 8:** If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

**Solution:**

Given: ΔABDC in which side BC is produced to D forming exterior angle ∠ACD of ΔABC.

To Prove: ∠4 = ∠1 + ∠2

Proof: Since, ∠1 + ∠2 + ∠3 = 180°…………equation (i) (angle sum of triangle)

∠2 + ∠4 = 180° ………….equation (ii) (Linear pair)

From equations (i) and (ii)

∠1 + ∠2 + ∠3 = ∠3 + ∠4

Or, ∠1 + ∠2 + ∠3 - ∠3 = ∠4

Or, ∠1 + ∠2 = ∠4

Hence, ∠4 = ∠1 + ∠2 Proved

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