Parallel Lines And A Transversal

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

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

Axiom 4 If a transversal intersects two lines such that a pair of matching angles is equal, then the two lines are parallel to each various other.

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

Theorem 2 If a transversal intersects two parallel lines, then each pair of alternative internal 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 = ∠SMLAnd ∠LMR = ∠MLQ

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Proof: ∠PLM = ∠RMB ………….equation (i) (Corresponding ngles)∠RMB = ∠SML ………….equation (ii) (vertically oppowebsite angles)From equation (i) and also (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 2 lines such that a pair of different interior angles is equal, then the two lines are parallel.

Solution: Given: - A transversal AB intersects 2 lines PQ and also 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 oppowebsite angles)From equations (i) and also (ii);∠PLM = ∠RMB

But these are equivalent angles.We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the 2 lines ate parallel to each various other.Hence, PQ║RS Proved.

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

Solution:Solution:

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Given: Transversal EF intersects 2 parallel lines AB and also CD at G and 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) (equivalent angles)From equations (i) and (ii),∠2 + ∠3 = 180°Also, ∠3 + ∠4 = 180° ………equation (iii) (Linear pair)But ∠3 = ∠1 …………..equation (iv) (Alternating internal angles)From equations (iii) and also (iv)∠1 + ∠4 = 180° and also ∠2 + ∠3 = 180° Proved

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

Solution:

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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) (Liclose to Pair)From equations (i) and (ii)∠1 + ∠2 = ∠1 + ∠3Or, ∠1 + ∠2 - ∠1 = ∠3Or, ∠2 = ∠3

But these are different interior angles. We understand that if a transversal intersects two lines such that the pair of alternative 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 various other.

Solution:

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

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

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

Angle Sum Property 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: 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 internal angles)∠2 = ∠5 ………..equation (ii) (alternative internal angles)By including equation (i) and (ii)∠1 + ∠2 = ∠4 + ∠5 ………..equation (iii)Now, by including ∠3 to both sides of equation (iii), we get∠1 + ∠2 + ∠3 = ∠4 + ∠5 + ∠3Since, ∠4 + ∠5 + ∠ = 180° (Liclose to group of angle)Hence, ∠1 + ∠2 + ∠3 = 180°Hence Proved.

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Theorem 8: If a side of a triangle is developed, then the exterior angle so created is equal to the amount of the 2 internal oppowebsite angles.

Solution:

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Given: ΔABDC in which side BC is produced to D creating exterior angle ∠ACD of ΔABC.To Prove: ∠4 = ∠1 + ∠2

Proof: Because, ∠1 + ∠2 + ∠3 = 180°…………equation (i) (angle sum of triangle)∠2 + ∠4 = 180° ………….equation (ii) (Liclose to pair)From equations (i) and (ii)∠1 + ∠2 + ∠3 = ∠3 + ∠4Or, ∠1 + ∠2 + ∠3 - ∠3 = ∠4Or, ∠1 + ∠2 = ∠4Hence, ∠4 = ∠1 + ∠2 Proved