Let $\alpha(t): I \to R^2$ be a regular parametrized curve. Assume that $k(t) \neq 0$. The evolute is defined as the curve: $$\beta(t)=\alpha(t)+\frac{1}{k(t)}n(t)$$ Consider the normal lines of $\alpha$ at two neighboring points $t_1,t_2$. Let $t_1$ approach $t_2$ and show that the intersection point converges.
I solved the problem and I want an opinion about it (if it is correct) and if you can find a more elegant solution. This one seems too much like brute force to me:
Consider the lines normal to the curve $\alpha$ at the points $s_1$ and $s_2$: $$d_1(\lambda)=\alpha(s_1)+\lambda n(s_1)$$ $$d_2(\mu)=\alpha(s_2)+\mu n(s_2)$$ We get the intersection point $P$ for $\lambda$ and $\mu$ that solve the equation: $$\alpha(s_1)+\lambda n(s_1)=\alpha(s_2)+\mu n(s_2)$$
Multiplying both sides with the tangent vector $t(s_2)$ (which is normal with $n(s_2)$) we get: $$\alpha(s_1)\cdot t(s_2)+\lambda n(s_1)\cdot t(s_2)=\alpha(s_2)\cdot t(s_2)$$ $$\lambda = \frac{(\alpha(s_1)-\alpha(s_2))\cdot t(s_2)}{n(s_1)\cdot t(s_2)}$$ Now I only need to prove that $\lambda = \frac{1}{k(s_1)}$ when $s_2 \to s_1$ and then replace it at $d_1$ to get the point $P$. $$\lim_{s_2 \to s_1}{\lambda}=\lim_{s_2 \to s_1}{\frac{(\alpha(s_1)-\alpha(s_2))\cdot t(s_2)}{n(s_1)\cdot t(s_2)-\underbrace{n(s_2)\cdot t(s_2)}_{=0}}}=$$ $$=\lim_{s_2 \to s_1}{\frac{\frac{\alpha(s_1)-\alpha(s_2)}{s_2-s_1}\cdot t(s_2)}{\frac{n(s_1)-n(s_2)}{s_2-s_1}\cdot t(s_2)}}=\lim_{s_2 \to s_1}{\frac{\alpha'(s_1)\cdot t(s_2)}{n'(s_1)\cdot t(s_2)}}=\frac{|t(s_1)|^2}{k(s_1)|t(s_1)|^2}=\frac{1}{k(s_1)}$$ as required.
