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Gaurav Seth 2 weeks ago

Theorem 6. 1 :Given :-To prove :-∠AOC = ∠BOD and ∠AOD = ∠BOCProof :- AB and CD be two lines intersecting at O as shown in Fig 6 8 They lead to two pairs ofvertically opposite angles, namely, (i) ∠AOC and ∠ BOD

(ii) ∠ AOD and ∠ BOC

We need to prove that ∠AOC = ∠BOD

and ∠AOD = ∠BOC

Now, ray OA stands on line CD

From (Linear pair axiom)

∠AOC + ∠AOD = 180° … (1)

∠AOD + ∠BOD = 180° … (2)

From (1) and (2), we can write

∠AOC + ∠AOD = ∠AOD + ∠BOD

This implies that ∠AOC = ∠BOD

Similarly, it can be proved that ∠AOD = ∠BOC

Parallel Lines and a Transversal∠1, ∠ 2, ∠7 and ∠ 8 are called

exterior angles,∠ 3, ∠ 4, ∠ 5 and ∠6 are called

interior angles(a)

Corresponding angles :(i) ∠1 = ∠ 5 (ii) ∠ 2 = ∠ 6 (iii) ∠ 4 = ∠ 8 (iv) ∠ 3 = ∠ 7

(b)

Alternate interior angles:(i) ∠ 4 = ∠ 6 (ii) ∠ 3 = ∠ 5

(c)

Alternate exterior angles:(i) ∠1 = ∠ 7 (ii) ∠ 2 = ∠8

(d)

Interior angles on the same side of the transversal:(i) ∠4 + ∠5 =180°(ii) ∠3 + ∠ 6 =180°

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