Rhombus Diagonal Properties: Analyzing Triangle Congruence in Four-Part Division

First, let's mark the vertices of the rhombus with the letters ABCD, then draw the diagonals AC and BD, and mark their intersection point with the letter E:

Now let's use several facts and properties:

a. The rhombus is a type of parallelogram, therefore its diagonals intersect each other, meaning:

$AE=EC=\frac{1}{2}AC\\ BE=ED=\frac{1}{2}BD\\$

b. A property of the rhombus is that its diagonals are perpendicular to each other, meaning:

$AC\perp BD\\ \updownarrow\\ \sphericalangle AEB=\sphericalangle BEC=\sphericalangle CED=\sphericalangle DEA=90\degree$

c. The definition of a rhombus - a quadrilateral where all sides are equal, meaning:

$AB=BC=CD=DA$

Therefore, from the three facts mentioned in: a-c and using the SAS (Side-Angle-Side) congruence theorem, we can conclude that:

d.
$$$\triangle AEB\cong\triangle CEB\cong\triangle AED\cong\triangle CED$(where we made sure to properly and accurately match the triangles according to their vertices in correspondence with the appropriate sides and angles).

Indeed, we found that the diagonals of the rhombus create (together with the rhombus's sides - which are equal to each other) four congruent triangles.

Therefore - the correct answer is answer a.