User:Rayyanahmed1423/Rayyan Theorem
THE PROOF IS ORIGINAL AND DONE BY RAYYAN AHMED OR WIKIPEDIA USERNAME - Rayyanahmed1423
THE PROOF IS INDEPENDENT OF ANY OTHER MAJOR THEOREM RELATED TO THE SUBJECT .
THE THEOREM : IF TWO LINES INTERSECT TWO PARALLEL LINES THEN THE INTERCEPTS ARE IN THE SAME RATIO
LETS DISCUSS MY PROOF
LETS SEE THE DIAGRAM FIRST
GIVEN ELEMENTS :-
<math>\angle A = \angle D
</math>
<math>\angle B = \angle E</math>
<math>\angle C = \angle F</math>
<math>OC = R , PF = r </math>
PROOF :-
SINCE , IF ARCS SUBTEND EQUAL ANGLES AT THE CENTRE OR CIRCUMFERENCE THEN THE ARCS ARE SIMILAR
<math>\left ( \frac{arcAC}{arcDF} \right )= \left ( \frac{\left ( \frac{2\angle B}{360 ^\circ} \right ) \times2\pi R}{\left ( \frac{2\angle E }{360 ^\circ} \right ) \times 2\pi r} \right ) = \left ( \frac{R}{r} \right ) </math> --------------- equation 1
<math>\left ( \frac{arcAB}{arcDE} \right )= \left ( \frac{\left ( \frac{2\angle C}{360 ^\circ} \right ) \times2\pi R}{\left ( \frac{2\angle F }{360 ^\circ} \right ) \times 2\pi r} \right ) = \left ( \frac{R}{r} \right ) </math>--------------- equation 2
<math>\left ( \frac{arcBC}{arcEF} \right )= \left ( \frac{\left ( \frac{2\angle A}{360 ^\circ} \right ) \times2\pi R}{\left ( \frac{2\angle D }{360 ^\circ} \right ) \times 2\pi r} \right ) = \left ( \frac{R}{r} \right ) </math>--------------- equation 3
AS FROM ABOVE EQUATIONS (1,2,3) , WE CAN CONCLUDE FROM EUCLID'S POSTULATES THAT "IF THINGS ARE EQUAL TO THE SAME OBJECT THEN THE OBJECTS SRE CONSIDERED TO BE EQUAL TO EACH OTHER "
FROM ABOVE :-
<math>\left ( \frac{arcAB}{arcDE} \right ) = \left ( \frac{R}{r} \right ) </math>
<math>\left ( \frac{arcBC}{arcEF} \right ) = \left ( \frac{R}{r} \right ) </math>
<math>\left ( \frac{arcAC}{arcDF} \right ) = \left ( \frac{R}{r} \right ) </math>
HENCE :- <math>\left ( \frac{arcAC}{arcDF} \right ) = </math><math>\left ( \frac{arcAB}{arcDE} \right ) = </math><math>\left ( \frac{arcBC}{arcEF} \right ) </math>
IF ALL CHORDS ARE SIMILAR THEN THE CORRESPONDING ARCS ARE ALSO SIMILAR
Therefore:- <math>\left ( \frac{AB}{DE} \right ) = \left ( \frac{AC}{DF} \right ) = \left ( \frac{BC}{EF} \right ) </math>
THE PROOF IS ORIGINAL AND DONE BY RAYYAN AHMED OR WIKIPEDIA USERNAME - Rayyanahmed1423