Lines and Angles

Parallel Lines And A Transversal

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

lines & angles 3

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

lines & angles 3

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:

lines & angles 9

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:

lines & angles 12

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:

lines & angles 13

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.

Angle Sum Property of Triangle:

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

Solution:

angle sum triangle

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:

external angle triangle

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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