Euclid's Proposition 1 and American Presidential Politics:
IN ORDER FOR EUCLID's PROPOSITION TO APPLY TO THE NEXT PRESIDENTIAL ELECTION IN 2016, A BLACK MAN (B) MUST FOLLOW BARACK OBAMA (A) INTO THE PRESIDENCY. THEREAFTER, A THIRD POINT (C) COULD BE LOCATED FROM THE INTERSECTION OF THEIR TWO DESCRIBED ORBITS. FROM THIS POINT (C), AN EQUILATERAL TRIANGLE CAN BE CONSTRUCTED, BASED UPON WHICH, FUTURE PROPOSITIONS CAN BE PROVEN AND RESOLVED. THE PROOF IS OUTLINED BELOW.
Let AB be the given finite straight line.
Describe the circle BCD with the center A and the radius AB.
Again describe the circle ACE, with the center B and the radius AB.
Join the straight line CA and CB from the point C at which point the circles cut one another to the points A and B.
Now since point A is the center of CDB, therefore, AC equals AB. Again, since point B is the center of CAE, therefore, CB equals BA.
But BC was proved equal to AB. Therefore each of the straight lines AC and BC are equal to AB.
And things which equal the same thing also equal one another. Therefore, AC and BC equals AB.
Therefore the straight lines AC , AB and BC equal one another.
Therefore the triangle ABC is equilateral, and has been connected on the given finite line AB.
IN ORDER FOR EUCLID's PROPOSITION TO APPLY TO THE NEXT PRESIDENTIAL ELECTION IN 2016, A BLACK MAN (B) MUST FOLLOW BARACK OBAMA (A) INTO THE PRESIDENCY. THEREAFTER, A THIRD POINT (C) COULD BE LOCATED FROM THE INTERSECTION OF THEIR TWO DESCRIBED ORBITS. FROM THIS POINT (C), AN EQUILATERAL TRIANGLE CAN BE CONSTRUCTED, BASED UPON WHICH, FUTURE PROPOSITIONS CAN BE PROVEN AND RESOLVED. THE PROOF IS OUTLINED BELOW.
Let AB be the given finite straight line.
Describe the circle BCD with the center A and the radius AB.
Again describe the circle ACE, with the center B and the radius AB.
Join the straight line CA and CB from the point C at which point the circles cut one another to the points A and B.
Now since point A is the center of CDB, therefore, AC equals AB. Again, since point B is the center of CAE, therefore, CB equals BA.
But BC was proved equal to AB. Therefore each of the straight lines AC and BC are equal to AB.
And things which equal the same thing also equal one another. Therefore, AC and BC equals AB.
Therefore the straight lines AC , AB and BC equal one another.
Therefore the triangle ABC is equilateral, and has been connected on the given finite line AB.
