 Welcome friends to another session of trigonometry. In this video, we are going to discuss a theorem which says the length of the circumference of a circle bears a constant ratio to its diameter. So all of us know that circumference to diameter of a circle is a very famous constant that's called pi. Circumference of any circle divided by the diameter of that circle is always a constant and that constant we know that as pi. So now we are going to prove that circumference to diameter ratio for any circle is constant. We will not be evaluating the value of pi but we will be proving that for any circle or for all circles the ratio of circumference to its diameter is always a constant. How do we prove it? So let us draw two concentric circles which is shown here in the figure. So there are two concentric circles with the common center O. Let us also inscribe a regular polygon. So if you see I have shown few of the sides of a regular polygon whose vertices are a dash, b dash, c dash, d dash and so on and so forth. This is a n sided regular polygon. Regular polygon means all the sides are equal. Regular polygon means all the sides are equal. Now let us say O A dash, O B dash, O C dash and O D dash are all radii of this outer circle. So O A dash, O B dash, O C dash and O D dash these are radii of outer circle. Similarly O A dash, O B dash, O C dash and O D dash cut the inner circle at point A, B, C and D. Now if that is so then O A, O B, O C and O D are radii of inner circle. Now in triangles let us say O AB, O AB and O AB and triangle O A dash, B dash. Clearly O A by O A dash is equal to O B by O B dash. Why? They are all equal to R by capital R where R is the radius of the inner circle and capital R is the radius of the outer circle. This hence, hence by converse of, converse of Thales theorem which you would have already studied Thales theorem or which is also called basic proportionality theorem. I am writing the acronym B P T, basic proportionality theorem. We can say that, what we can say is AB is parallel to, if you see AB is parallel to A dash, B dash. This is coming from the reverse, oh sorry the converse of basic proportionality theorem. So we can say AB is parallel to A dash, B dash, A dash, B dash. Similarly, if you see, similarly BC is parallel to B dash, C dash and others like CD is parallel to C dash, B dash and so on and so forth. So all these are parallel. So AB is parallel to A dash, B dash like that. Okay now when they are parallel, so what we can say about these, let me redraw these two triangles. So this is, let us say this was our triangle and yeah, so this is O, this is A, this is B, this is C, sorry A dash and this is B dash. And AB is parallel to A dash, B dash, that means this angle is equal to this angle and this angle is equal to that angle. Now again in triangle OAB and triangle OA dash, B dash, what do we observe? We observe that angle OAB is equal to angle OA dash, B dash because AB is parallel to A dash, B dash. Similarly, angle OBA is equal to angle OB dash, A dash therefore by AA similarity criteria, angle angle similarity criteria, we can say triangle OAB is similar to triangle OA dash, B dash. Hence when two triangles are similar, what happens? The ratio of corresponding sides are equal. So hence we can say OA by OA dash is equal to AB by A dash, B dash and this is equal to small r by capital R. Why small r by capital R? Because OA is equal to small r and OA dash is equal to capital R, right? Radius of the two circles. Now let us find out the perimeter of the two polygons now. Hence I am saying perimeter of inner polygon upon perimeter of outer polygon. I will tell you why I am doing this but let us calculate this. So perimeter of inner polygon upon perimeter of outer polygon is nothing but N times N is the side and one side of inner polygon was AB. This upon N times A dash, B dash. So perimeter of inner polygon divided by perimeter of outer polygon is nothing but N times smaller side AB and N times the A dash, B dash which is the side of the outer polygon. This is AB, so N times AB, N sided polygon it was. So N times AB and N times A dash, B dash. This is A dash, B dash and this is AB, right? This is what the ratio of perimeter of the two polygon. Okay, so hence which is nothing but AB upon A dash, B dash. And we have just proved above here that AB by A dash, B dash is equal to small r by capital R, okay? By similarity we just proved above. Now when N tends to infinity, right? Polygon, the regular polygon tends to circle, the shape will convert, get converted into circle. What can we say about the perimeter we can say? As N tends to infinity, circumference of inner circle upon circumference, circumference of outer circle, outer circle is equal to small r by capital R. Hence, hence, circumference, circumference of inner circle by small r is equal to circumference of outer circle by capital R, right? So if you divide both sides by two, what will you get? You will get, dividing both sides by two, you will get circumference of inner circle divided by two r is equal to circumference of outer circle by two capital R. And what is two r? Simply D diameter. So hence I can say, circumference, circumference of inner circle, inner circle, divided by the diameter of inner circle, diameter of inner circle is equal to circumference of outer circle, circumference of outer circle divided by dia of outer circle, right? Now this ratio doesn't depend on the size of the circle, that is the radius of the circle. So hence it is true for all circles. So hence we say the ratio of circumference of inner circle upon the dia of inner circle is a constant. That is what we had an objective to prove, right? Whether it is any value of r, this relation will always hold. So hence what do we learn? We learn that circumference of any circle divided by diameter of that circle is always a constant value, which we now know is pi.