Theorem on Alternate Angles Axiom

Theorem regarding alternate angles
If two parallel lines are intersected by a transversal then the two pairs of alternate angles are equal. We shall prove this theorem by using the axiom of corresponding angles.
We have to prove that a) <4 = <6 and b) <3 = <5
The pair of angles formed on a stright line are called linear pair and the sum of these angles will be equal to 180 degrees.
In the above figure the following sets of linear pair angles a) <2 and <3 b) <5 and <6. The sum of these angles is 180 degrees.
<2 + <3 = 180 and <5 + <6 + 180. In other words we can say that <2 + <3 = <5 + <6. Since <2 and <6 are corresponding angles they are equal and we can cancel out these equal angles from both sides and finally we arrive at the conclusion that <3 = <5.
The same method is used to prove that <4 = <6.

Uses of Alternate Angles Axiom

We shall use this theorem on alternate angles to prove that the sum of the three angles of a triangle is equal to 180 degrees. "l" and "m" are two parallel lines in a plane in between which we have a triangle ABC as shown below. Angles are marked as 1,2,3,4 and 5 as shown in the figure.
The angle sum property of triangle states that the sum of the 3 angles of a triangle = 180 degrees. We shall prove this concept by using the theorem on alternate angles. In the figure we have to prove that <4 + <2 + <5 = 180 degrees. We know that angle in a striaght line is 180 degrees. Since "l" is a straight line we can write <1+<2+<3=180. But <1=<4 and <3=<5. Hence we can write that <4+<2+<5=180 degrees. Thus by using the concept of alternate angles we have easily proved that the sum of 3 angles of a traingle=180 degrees.


Principle of Duality

  • Any theorem or identity in a switching algebra will be true if 0 and 1 are swap and . and + are also be swapped.
  • If it becomes True because of all the duals of all the axioms are true, so duals of all switching-algebra theorems can be proved by using the duals of axioms.
  • Dual of a logic expression
The Principle of dual expression is compensated with fully parenthesized logic expression FD( A1, A2, …., An, +, . , ’), FD, is the same expression with +, . they will be swapped:
  • After swapped

FD( A1, A2, …., An, +, . , ’) = F(A1, A2, …., An, . , + , ’)
  • The generalized DeMorgan’s theorem T14 can be stated as

[F( A1, A2, …., An)]’ = FD(A1’, A2’, …., An’)

Primal and Duality Problems Relationship:-

In the primal problem, from each sub-optimal point that gratifies all the constraints, there is a track or subspace of orders to move that boost the objective function.
In the dual problem, the dual vector multiplies the constants that conclude the positions of the constraints in the primal. Altering the dual vector in the dual problem is corresponding to revising the higher bounds in the primal problem
Hosted by Free Image Hosting Service
Hosted by Free Image Hosting Service