If a graph of $2n$ vertices contains a Hamiltonian cycle, then can we reach every other vertex in $n$ steps? 
Problem: Given a graph $G,$ with $2n$ vertices and at least one triangle. Is it possible to show that you can reach every other vertex in $n$ steps if $G$ contains a Hamilton cycle (HC)?
EDIT: Sorry, I forgot to mention, that $G$ is planar and 3-connected. A complete proof for $3$-regular graphs would also be accepted/rewarded.

Does the following work as proof?

Choose a starting vertex $v_0$ and a direction.

*

*If you walk along the HC you'll reach a vertex $v_{n-1}$ with maximal distance from $v_0$ in $n$ steps.

*You'll reach $v_{n-2}$ by doing a round in the triangle and

*$v_{n-3}$ by stepping backwards at the last step.

*By combining these moves, you'll reach half of all $v_k$.

*By choosing the other direction at the beginning you'll reach the other half.

*$v_0$ is free to choose.


Showing or disproving the "only if"-part would also be nice!

The answer is no.
Question: Let $G$ be a 3-connected, hamiltonian, planar graph with $2n$ vertices and at least one triangle. Is it true that for all vertex pairs $x,y$, that there is a walk of exactly $n$ steps from $x$ to $y$?
The following graph and vertex pair is a counter example

It is clear that the graph is planar and has a triangle. It can be easily verified that the graph is 3-connected. To show that the graph is hamiltonian, I have highlighted a hamiltonian cycle here

Since the vertex has 16 vertices, we need to verify that there is no walk of length 8 from $x$ to $y$. Since $n$ is equal, we can not reach $y$ without using some of the four vertices on the right. Now it is easy to verify by hand, that there is no walk from $x$ to $y$ of length exactly 8.
