I asked a question a few days ago and figured out the proof for this theorem using induction.
If T is a full binary tree with i internal vertices, then T has i + 1 terminal vertices and 2i + 1 total vertices.
Now as a challenge I am trying to use the Handshaking Lemma to show the same thing.
H.S. Lemma: Every finite undirected graph has an even number of vertices with odd degree.
I have this written out but am stuck on how to continue.
Each vertex of a FBT has 0 or 2 children, if the child is a terminal vertex it has a degree of 1 (to its parent) and if the child is a internal vertex it has a degree of 3 (to its parent and two children). The only non-child is the root node. Because all internal vertices must have 2 children there exists 2i children. A child whether internal or terminal has a odd degree (1 or 3). Thus there are an even number of vertices 2i with odd degree.