Editing Brachistochrone curve (section)

==== Synthetic solution ====
He then proceeds with what he called his Synthetic Solution, which was a classical, geometrical proof, that there is only a single curve that a body can slide down in the minimum time, and that curve is the cycloid.

"The reason for the synthetic demonstration, in the manner of the ancients, is to convince [[Philippe de La Hire|Mr. de la Hire]]. He has little time for our new analysis, describing it as false (He claims he has found 3 ways to prove that the curve is a cubic parabola)" – Letter from Johan Bernoulli to Pierre Varignon dated 27 Jul 1697. <ref>{{cite book |last1= Costabel |first1= Pierre| first2= Jeanne|last2= Peiffer| title= "Der Briefwechsel von Johann I Bernoulli", Vol. II: "Der Briefwechsel mit Pierre Varignon, Erster Teil: 1692-1702"|date=1988 |publisher= Springer Basel Aktiengesellschaft |isbn=978-3-0348-5068-1 |pages=117–118|edition=Hardback}}</ref>

Assume AMmB is the part of the cycloid joining A to B, which the body slides down in the minimum time. Let ICcJ be part of a different curve joining A to B, which can be closer to AL than AMmB.  If the arc Mm subtends the angle MKm at its centre of curvature, K, let the arc on IJ that subtends the same angle be Cc. The circular arc through C with centre K is Ce. Point D on AL is vertically above M. Join K to D and point H is where CG intersects KD, extended if necessary.

Let <math> \tau </math> and t be the times the body takes to fall along Mm and Ce respectively.

:<math> \tau \propto \frac{Mm}{MD^{\frac{1}{2}}} </math>, <math> t \propto \frac{Ce}{CG^{\frac{1}{2}}} </math>, 
Extend CG to point F where, <math> CF = \frac{CH^2}{MD} </math> and since <math>  \frac{Mm}{Ce}  = \frac{MD}{CH} </math>, it follows that
:<math> \frac {\tau}{t} = \frac{Mm}{Ce}.\left({\frac{CG}{MD}}\right)^{\frac{1}{2}} = \left({\frac{CG}{CF}}\right)^{\frac{1}{2}} </math>
Since MN = NK, for the cycloid:
:<math> GH = \frac{MD.HD}{DK} = \frac{MD.CM}{MK} </math>, <math> CH = \frac{MD.CK}{MK} = \frac{MD.(MK + CM)}{MK} </math>, and <math> CG = CH + GH = \frac{MD.(MK + 2CM)}{MK} </math>

If Ce is closer to K than Mm then 
:<math> CH = \frac{MD.(MK - CM)}{MK} </math> and <math> CG = CH - GH = \frac{MD.(MK - 2CM)}{MK} </math>
In either case, 
:<math> CF = \frac{CH^2}{MD} > CG </math>, and it follows that <math> \tau < t </math>
If the arc, Cc subtended by the angle infinitesimal angle MKm on IJ is not circular, it must be greater than Ce, since Cec becomes a right-triangle in the limit as angle MKm approaches zero.

Note, Bernoulli proves that CF > CG by a similar but different argument.

From this he concludes that a body traverses the cycloid AMB in less time than any other curve ACB.