Show that there exists $d=(1-t)a+tc$ with $t\in(0,1)$ such that $||b-a||=||b-d||$

Question

I have a problem that I could solve if I could demonstrate the following.

Let $a,b,c\in\mathbb{R}^n$ such that $||b-a||<||b-c||$ and suppose that $\langle b-a, c-a\rangle>0$. Show that there exists $d=(1-t)a+tc$ with $t\in(0,1)$ such that $||b-a||=||b-d||$.

Geometrically the problem is very obvious and in fact I was able to solve it using arguments from Euclidean geometry. I would like to solve it using properties of the inner product and the norm if I could.

Answer 1

Define $f(t):=\|b-a\|^2-\|b-d(t)\|^2=\langle b-a,b-a\rangle-\langle b-d(t),b-d(t) \rangle.$ Then since $d(0)=a$, clearly $f(0)=0.$ From the hypothesis $\|b-a\|<\|b-c\|$, we also have (since $d(1)=c$) that $f(1)<0$.

$f'(t)=-\left(\langle -d'(t),b-d(t)\rangle+\langle b-d(t),-d'(t)\rangle\right)=2\langle d'(t),b-d(t)\rangle.$

$d'(t)=c-a$ so $f'(0)=2\langle c-a,b-a\rangle>0$ by hypothesis.

It is obvious from the intermediate value theorem that there exists $t_0\in(0,1)$ such that $f(t_0)=0$.