minimal number of vertices of a graph that can be split into two trees

Question

Suppose a graph has no 3 cycle, and can be split into two trees, that is the edges can be partitioned into two sets, each part form a tree. For example: (red and green indicate partition)

I want to show that such graphs have at least $k-1$ vertices, where $k$ is the number of edges. Could anyone give a hint, or potentially a counter example? The only intuition I have for this problem is that the graph probably look like something in the image.

Answer 1

Consider $K_{n,2}$ which has no $3$-cycle, since it's bipartite. The edges can be partitioned to form two copies of $K_{n,1}$. It has $n+2$ vertices and $2n$ edges.

Answer 2

counter example

This graph has 6 vertices and 8 edges and contains no 3-cycles. It can be split in two trees in the same way that the original with 5 vertices and 6 edges can.