Friendship Theorem/Proof 2

Proof
This can be expressed in graph theoretical language as:

Let the edges of $K_6$, the complete graph of $6$ vertices be edge-colored using $2$ colors.

Then there will be a triangle subgraph of $K_6$ whose edges all have the same color.

Let the colors be red and blue.

Choose any one vertex $P$.

There are $5$ edges incident to $P$.

They are each colored red or blue.

By the Pigeonhole Principle, at least $3$ of them must be of the same color.

, suppose that less than $3$ are red.

Then the others, $3$ or more, must be blue.

Let $A$, $B$, $C$ be the other vertices incident to these $3$ edges, all of which are blue.

There are $2$ cases:


 * $(1): \quad$ Let any one of $AB$, $BC$ and $CA$ be blue.

Then that edge, together with the $2$ edges to $P$ forms a blue triangle.


 * $(2): \quad$ Let none of $AB$, $BC$ and $CA$ be blue.

Then all three edges are red.

Thus $AB$, $BC$ and $CA$ form a red triangle.

Hence the result.