Mental block with geometry?

Question

So I've tried playing with this question but I can't seem to get anywhere...

A rhombus is a convex quadrilateral all of whose edges are equal in length. Given a rhombus ABCD and a point O prove that the distance from O to one of the vertices of the rhombus is not greater than the sum of the distances from O to the other vertices.

Any ideas?

Answer 1

make 4 triangles by connecting O to each of 4 vertices.

then somehow try to use the fact that sum of two sides of triangle is larger than the 3d side. you have 4 of these inequalities, and some sides are equal, so maybe you can chain the inequalities in some way.

Answer 2

Suppose that the rhombus were really flat. by that I mean two of the vertexes are almost coincident and the other two are wide apart. If O is on one of the wide apart vertexes then the other wide apart vertex is the farthest away from the point O. If it can be proved that O is not farther from that vertex than the sum of the distances to the other three vertexes then the distance to none of the other vertexes could be larger either.

The distance from O to the vertex upon which it sits is zero.
The distance to each of the almost coincident vertexes is equal to the length of the side of the rhombus.
The sum of these three distances is then equal to twice the length of the side of the rhombus.
The distance to the far vertex is a straight line while the distance along the periphery of the rhombus is not. The distance to the far vertex is therefore something less than twice the length of the side of the rhombus since the shortest distance between two points is a straight line.
Moving O away from the vertex would need to be investigated. There are three cases of such movement.

Then describing similar analysis for rhombus that have the two almost coincident vertexes farther apart.