# Prove without calculus: 2 tangent segments to convex curve longer than curve

Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.

I would like to show, without calculus, that $$AB + AC > BC$$.

With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $$B'C'$$ below A, the slope of is zero. Call it D. In the diagram,

$$AB' = \int_y^z \sqrt{1+slope(AB')^2} dx$$

$$B'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$$

$$AC' = \int_y^z \sqrt{1+slope(AC')^2} dx$$

$$C'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$$

The absolute value of the slope of $$B'C'$$ is decreasing from the slope of $$AB'$$ from B' to D, and negative but increasing (up to the absolute value of the slope of $$AC'$$) from D to C'

This is a general case of this question, which helped me when trying to place an upper limit on Pi.

• I don't see how you can avoid calculus when you talk about tangent segments and curve lengths. – TonyK Jan 31 at 23:34