Prove four vertices of pentagon are concyclic?

First make sure you understand that concyclic simply means the points lie on a common circle. We are not told it is a regular pentagon but we will assume it is. We could create pentagons that are not even convex and would not be concyclic.

Let's start with a regular pentagon. You can split it up into 5 congruent triangles with the points meeting at the middle. Any side of one of these triangle is connected from each of the vertices of the pentagon to the center of the pentagon. Since all 5 triangles are congruent, this distance must be the same for each of the vertices. So, we see that each of the vertices is equidistant from a given point. Now if we drew a circle centered at that point with a radius equal to the distance between the point and any vertex, that circle would pass through all 5 vertices. Therefore any four ( really all 5), vertices of a regular pentagon are concyclic.

A nice proof would use Ptolemy's theorem. I will place a line to an answers.com page that helps with that.

Another solution:

The pentagon has to be regular. Otherwise, the question is impossible to prove.

Consider a regular polygon ABCDE.

Take triangles BCD & ECD

BC=ED (sides of a regular polygon are equal)

CD=CD (common side)

(interior angles of a regular polygon are equal)

Therefore,

Triangle BCD is congruent to triangle ECD

Therefore,

(corresponding parts of congruent triangles are equal)

Since these angles are equal, it proves the theorem "Equal chords of a circle subtend equal angles".

Thus, vertices B , C , D , E lie on a circle and are hence CONCYCLIC.