if s is arc length..then A is parametrized by arc length, means |A'(s)| = 1 for all s in I...

squaring it.. we get,

|A'(s)|.|A'(s)|=1.

differentiate it..

we'll get ... A'(s).A"(s)=0.. tht is tangent is perpendicular to a line..with slope A''(s)..

now take A as a line segment. so, all tangents to it will lie on itself at all points..,,that makes them concurrent.. (but i need to show that slope is constant) Can you guide me here..

For converse part ,im able to visualize n see how it looks..but i need to write it mathematically..

"if all tangents are concurrent then draw two tangents sharing a common point to a curve... now draw a third tangent to it,cutting both tangents at two points... now if all tangents are concurrent..the third shud coincide for all points..with either of 1st two..

or bring..contact points..on the curve at a minimum distance to each other..slope will reduce...and normals will coincide wen very near...tht portion of curve will hav almost a constant slope...or u can say it is a line segment of a straight line.."