Find different $f_1,f_2 \in F$ such that $\lVert f_1 \rVert = \lVert f_2 \rVert=1$ and $\lVert f_1 + f_2 \rVert =2$ 
I have asked this question before, but i did not get any satisfactory answer, so ill try again because i can't give up until I understand/derive this myself. Given a normed linear space $F$ equipped with a norm that is not strictly convex, are we garanteed to find different $f_1,f_2 \in F$ such that $\lVert f_1 \rVert = \lVert f_2 \rVert=1$ and $\lVert f_1 + f_2 \rVert =2$ ? (ofc assuming that  $F$ is not the space with only the zero element).

What do i mean (my professor) with a non strictly convex norm? I mean that the set $B=\{f \in F :  \lVert f \rVert \leq1 \} $ is not stricly convex. By the answer i got from my previous question, this seems to be always possible, but i can't prove it myself. Here is my approach.
Since the norm is not strictly convex, it exists different $f_1,f_2 \in B \subset F$ and some $t \in (0,1)$ such that $tf_1+(1-t)f_2$ is contained in the boundary of $B$. With other words, there exists a $t$ such that $\lVert tf_1+(1-t)f_2\rVert =1$ where $\lVert f_1 \rVert , \lVert f_2 \rVert \leq1$.
Am I correct so far?
In the answer though, it says that we can always find different $f_1,f_2$ such that $\lVert f_1 \rVert = \lVert f_2 \rVert =1$ and a $t \in (0,1)$ such that $\lVert tf_1+(1-t)f_2\rVert =1$ Is this accurate? I fail to see what garantees that we can find such $f_1$ and $f_2$ on the boundary of the ball satisfying this conditions? Geometrically it makes sense when drawing different convex sets in the plane, so intuitively i feel this is true.

Now assuming all above is true, we have different $f_1 , f_2$ and a $t \in (0,1)$ so that $\lVert f_1 \rVert = \lVert f_2 \rVert =1$ and $\lVert tf_1+(1-t)f_2\rVert =1$. Is it true then that this implies that $\lVert \lambda f_1+(1-\lambda)f_2\rVert =1$ for all $\lambda \in [0,1]$ ?

If yes, how? I completely fail to prove that statement. Any help/tips appreciated, thanks.  
 A: If the norm is not strictly convex then the boundary of the unit ball contains a line segment, so there are distinct $x,y$ such that $\|t x + (1-t)y\| = 1$ for all $t \in [0,1]$.
Hence $\|{1\over 2} (x+y)\| = 1$, and so $\|x+y\| = 2$.
A: Proof that the line segment is in the unit sphere. Lets use $f_0,f_1$ instead of $f_1,f_2$, and write $f_t :=  tf_0 + (1-t)f_1$. The key observation is the following: if either $\|f_0\|$ or $\|f_1\|<1$, then observe that for $t\neq 0,1$,
$$\| f_t \| \leq t\|f_0\| + (1-t) \|f_1\| < 1.$$
Hence, $\|f_0\|\leq 1, \ \|f_1\|\leq 1, \ \|f_t\| = 1$  means that actually $\|f_0\| = \|f_1\| = 1$.
A similar argument works for the entire line segment as follows. Suppose $s>t$. Then $f_t$ is also a convex combination of $f_0$ and $f_s$,
$$ f_t = u f_0 + (1-u)f_s,\quad t = u + (1-u) s.$$ 
Similarly to the above, if $\|f_s\|<1$, then
$$ \|f_t\| \leq u \|f_0\| + (1-u)\|f_s\| < 1. $$
Hence $\|f_s\|=1$. A similar strategy works for $s<t$. Therefore the whole line $\{ f_t : t\in [0,1] \}$ is in the unit sphere, and $f_{1/2}$ is useful as in copper.hat's answer or your previous question.
