Sufficient Conditions for a Rhombus
Perpendicular diagonals, an angle-bisecting diagonal or two equal neighbouring sides each prove a parallelogram is a rhombus. These converses turn properties into tests.
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Perpendicular diagonals, an angle-bisecting diagonal or two equal neighbouring sides each prove a parallelogram is a rhombus. These converses turn properties into tests.
A rhombus has perpendicular diagonals, angle-bisecting diagonals and four equal sides. Each of those properties also runs the other way: spot any one of them in a parallelogram and the shape must be a rhombus. These are the converses of the theorems you already know.
If , then is a rhombus.
Note the starting point: the shape must already be a parallelogram. Perpendicular diagonals alone, in an arbitrary quadrilateral, only give a kite.
In , if the diagonal splits each corner angle into two equal halves, then is a rhombus.
One diagonal is enough. Once it bisects its two corner angles, the other diagonal is forced to do the same.
If , then is a rhombus.
Two sides settle all four. Opposite sides of a parallelogram are already equal, so making one neighbouring pair equal forces every side to the same length.
A quadrilateral that is both a rectangle and a rhombus is a square.
The two statements say the same thing from opposite ends. A rectangle supplies the right angles, a rhombus supplies the equal sides, and a square is the shape that has both.
In parallelogram the diagonals cross at , with and . Is it a rhombus?
The condition either holds throughout or not at all. Finding a value of that satisfies one angle proves nothing on its own — it has to satisfy the other as well.