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This cannot be proven, because it is not generally true. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. And conversely, the diagonals of any parallelogram bisect each other. However not every parallelogram is a rhombus.

However, if the diagonals are perpendicular bisectors, then we have a rhombus.

Consider quadrilateral ABCD, with diagonals intersecting at X, where

AC and BD are perpendicular;

AX=XC;

BX=XD.

Then angles AXB, BXC, CXD, DXA are all right angles and are congruent.

By the ASA theorem, triangles AXB, BXC, CXD and DXA are all congruent.

This means that AB=BC=CD=DA.

Since the sides of the quadrilateral ABCD are congruent, it is a rhombus.

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