# Decomposing planar Hamiltonian graphs

I have the following the statement and I have to prove whether it is true or not.

Given a planar and Hamiltonian graph $$\mathbb{G} = (V, E)$$, show that we can partition $$E = E_1 \cup E_2 = E$$ so that $$G_1 = (V, E_1)$$ and $$G_2 = (V, E_2)$$ are both connected and outerplanar graphs.

I have tested the statement for few graphs and it seems correct to me, but I am not sure how can I use the property of the Hamiltonian cycle to decompose the graph into two sub-graphs.

I appreciate any hint!

• What is the connection between $\mathbb{V},\mathbb{E},V,E_1,E_2$? – Yuval Filmus Nov 3 '20 at 14:20
• Sorry, I have just edited the post – Mohbenay Nov 3 '20 at 14:27
• Can you find a connected non-Hamiltonian planar graph for which this fails? – Yuval Filmus Nov 3 '20 at 14:43