theorem of chap.6

Theorem 6. 1 : If two lines intersect each other, then the vertically opposite angles are equal
Given :- two lines intersect each other
To prove :-   ∠AOC =   ∠BOD  and ∠AOD =  ∠BOC

Proof  :- AB and CD be two lines intersecting at O as shown in Fig  6 8  They lead to two pairs of
vertically 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°