Prove that if a transversal intersects two parallel lines, then alternate interior angles are equal

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 ∠8Interior angles are ∠3, ∠4, ∠5 and ∠6Corresponding 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 = ∠7Alternate Interior Angles: (i) ∠4 and ∠6 and (ii) ∠3 and ∠5Alternate Exterior Angles: (i) ∠1 and ∠7 and (ii) ∠2 and ∠8If 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 = ∠5Alternate 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 ||RSUse 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 + ∠3Or, ∠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.

Angle Sum Property of Triangle:

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 + ∠3Since, ∠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 + ∠4Or, ∠1 + ∠2 + ∠3 - ∠3 = ∠4Or, ∠1 + ∠2 = ∠4

Hence, ∠4 = ∠1 + ∠2 Proved

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Last updated at June 29, 2018 by Teachoo

Theorem 6.3 :- If a transversal intersects two lines, such that a pair of alternate interior angles is equal, then the two lines are parallel. Given :- Two lines AB and CD. And transversal PS intersecting AB at Q and CD at R, Such that alternate interior angles are equal. i.e, BQR = CRQ To Prove :- AB CD Proof :- For lines AB & PS AQP = BQR But, BQR = CRQ From (1) & (2), AQP = CRQ But they are corresponding angles. Thus, for lines AB & CD with transversal PS, corresponding angles are equal Hence AB and CD are parallel. Hence, proved