Now we're introducing "Limits" in Geometry!

Without loss of generality, we may assume that the side length $c = 1$ . Let us embed the triangle $ABC$ into the Euclidean plane $\mathbb{R}^2$ as follows. Place it so that the vertex $A$ is at the origin, the vertex $B$ at the point $(0,1)$ , and the point $C$ as a variable point $(c,0)$ . We then have explicit formulae for $w,m, h$ as functions of $c$ . In particular, we realize that $m(c)$ is simply half the length of the hypotenuse, whence $m(c) = \frac{\sqrt{1+c^2}}{2}$ . Next, let $W$ be the point of intersection of the hypotenuse and the line $y = x$ . Then $w(c)$ is the length of the line segment $\overline{AW}$ . The hypotenuse lies on the line with equation $y = \frac{-x}{c} + 1$ , whence $W$ has coordinates $\left(\frac{c}{c+1}, \frac{c}{c+1} \right)$ . Thus $w(c) = \frac{\sqrt{2} c}{c+1}$ . Finally, we may compute $h(c)$ via similar triangles and deduce that $\frac{h(c)}{1} = \frac{c}{\sqrt{1+c^2}}.$ Thus the desired limit is equal to $\lim_{c \rightarrow 1} \dfrac{\dfrac{\sqrt{1+c^2}}{2} - \dfrac{c}{\sqrt{1 + c^2}}}{\dfrac{\sqrt{2} c}{c+1} - \dfrac{c}{\sqrt{c^2+1}}} = \lim_{c \rightarrow 1} \frac{(c-1)^2 (c+1)}{2c(\sqrt{2c^2 + 2} - c - 1)}.$ We can evaluate this limit by the usual conjugation trick in first year calculus, to obtain the limit $\lim_{c \rightarrow 1} \frac{(c-1)^2 (c+1) (\sqrt{2c^2 + 2} + c + 1)}{2c(2c^2 + 2 - c^2 - 2c - 1)} = \lim_{c \rightarrow 1} \frac{(c+1)(\sqrt{2c^2 + 2} + c + 1)}{2c} = 4,$ by inspection.

<DAE=B-45°

w=h sec(B-45°)

<DAM = 2B - 90°

m = h sec(2B-90°)

b->c => B->45°

Put B-45°=X, the limit is reduced to

lim(X->0) ({sec 2X-1}/{sec X-1}) =4

Take A as origin, B as ( c,0) and C as (0,b). Now find equation of lineBC with 2-point form. Write length of perpendicular . To find length of median apply distance formula for origin and (b/2,c/2) . To find length of angle bisector find point of intersection of BC and line y=x. Now put the values in limit and rationalise to get answer as 4.