Parallel Lines and also A Transversal

Axiom 3: If a transversal intersects 2 parallel lines, then each pair of corresponding angles is equal.

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Here, Exterior angles room ∠1, ∠2, ∠7 and ∠8Interior angles room ∠3, ∠4, ∠5 and also ∠6Corresponding angles space ∠(i)∠1 and ∠5(ii)∠2 and ∠6(iii)∠4 and ∠8(iv)∠3 and ∠7

Axiom 4 If a transversal intersects 2 lines such the a pair of corresponding angles is equal, climate the two lines are parallel to each other.

Thus, (i) ∠1 = ∠5, (ii) ∠2 = ∠6, (iii) ∠4 = ∠8 and also (iv) ∠3 = ∠7Alternate interior Angles: (i) ∠4 and ∠6 and (ii) ∠3 and ∠5Alternate Exterior Angles: (i) ∠1 and also ∠7 and also (ii) ∠2 and ∠8If a transversal intersects 2 parallel lines then each pair of alternative interior and also exterior angles room equal.Alternate inner Angles: (i) ∠4 = ∠6 and also (ii) ∠3 = ∠5Alternate Exterior Angles: (i) ∠1 = ∠7 and also (ii) ∠2 = ∠8Interior angles on the exact same side of the transversal line are referred to as the consecutive interior angles or allied angle or co-interior angles. They room as follows: (i) ∠4 and ∠5, and also (ii) ∠3 and also ∠6

Theorem 2 If a transversal intersects two parallel lines, climate each pair of alternating interior angles is equal.

Solution: Given: allow PQ and RS are two parallel present and abdominal be the transversal which intersects them on L and M respectively.

To Prove: ∠PLM = ∠SMLAnd ∠LMR = ∠MLQ

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Proof: ∠PLM = ∠RMB ………….equation (i) (Corresponding ngles)∠RMB = ∠SML ………….equation (ii) (vertically the contrary 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 alternating interior angles is equal, then the two lines space parallel.

Solution: Given: - A transversal ab intersects two lines PQ and also RS together that∠PLM = ∠SML

To Prove: PQ ||RSUse same number as in organize 2.Proof: ∠PLM = ∠SML ……………equation (i) (Given)∠SML = ∠RMB …………equation (ii) (vertically opposite angles)From equations (i) and (ii);∠PLM = ∠RMB

But this are matching angles.We recognize that if a transversal intersects 2 lines such that a pair of equivalent angles is equal, then the 2 lines ate parallel to every other.Hence, PQ║RS Proved.

Theorem 4: If a transversal intersects two parallel lines, climate each pair of inner angles on the exact same side of the transversal is supplementary.

Solution:Solution:

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Given: Transversal EF intersects 2 parallel lines ab and CD in ~ G and also H respectively.To Prove: ∠1 + ∠4 = 180° and also ∠2 + ∠3 = 180°

Proof: ∠2 + ∠5 = 180° ………equation (i) (Linear pair of angles)But ∠5 = ∠3 ……………equation (ii) (corresponding angles)From equations (i) and also (ii),∠2 + ∠3 = 180°Also, ∠3 + ∠4 = 180° ………equation (iii) (Linear pair)But ∠3 = ∠1 …………..equation (iv) (Alternate internal angles)From equations (iii) and also (iv)∠1 + ∠4 = 180° and also ∠2 + ∠3 = 180° Proved

Theorem 5: If a transversal intersects 2 lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Solution:

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Given: A transversal EF intersects 2 lines abdominal muscle and CD in ~ P and also 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 = ∠3Or, ∠2 = ∠3

But these are alternate interior angles. We know that if a transversal intersects 2 lines such that the pair of alternative interior angles space equal, climate the lines room parallel.Hence, AB║CD Proved.

Theorem 6: lines which are parallel come the exact same line space parallel to each other.

Solution:

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

∠1 = ∠2 ………….equation (i) (corresponding angles)Similarly, CD ||EF and also GH is transversal. Therefore;∠2 = ∠3 ……………equation (ii) (corresponding angles)From equations (i) and also (ii)∠1 = ∠3

But this are equivalent angles.We understand that if a transversal intersects 2 lines such the a pair of equivalent angles is equal, then the two lines ate parallel to each other.Hence, AB║ EF Proved.

Angle Sum building of Triangle:

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

Solution:

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Given: Δ ABC.To Prove: ∠1 + ∠2 + ∠3 = 180°Construction: allow us draw a line m though A, parallel to BC.

Proof: BC ||m and abdominal and AC are its transversal.Hence, ∠1 = ∠4 …………….equation (i) (alternate interior angles)∠2 = ∠5 ………..equation (ii) (alternate interior angles)By including equation (i) and also (ii)∠1 + ∠2 = ∠4 + ∠5 ………..equation (iii)Now, by including ∠3 come both sides of equation (iii), us get∠1 + ∠2 + ∠3 = ∠4 + ∠5 + ∠3Since, ∠4 + ∠5 + ∠ = 180° (Linear group of angle)Hence, ∠1 + ∠2 + ∠3 = 180°Hence Proved.

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Theorem 8: If a side of a triangle is produced, climate the exterior angle so created is same to the sum of the two internal opposite angles.

Solution:

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Given: ΔABDC in which next BC is created to D forming exterior edge ∠ACD that ΔABC.To Prove: ∠4 = ∠1 + ∠2

Proof: Since, ∠1 + ∠2 + ∠3 = 180°…………equation (i) (angle amount 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 = ∠4Hence, ∠4 = ∠1 + ∠2 Proved