The rhombus and the perpendicular diagonals

A parallelogram becomes a rhombus (all sides congruent) as soon as its diagonals are perpendicular. This is proved with Pythagoras’ theorem applied to the four right triangles into which the diagonals divide the figure.

Theorem — Rhombus (perpendicular diagonals)

If a parallelogram has perpendicular diagonals, then it is a rhombus (all sides congruent).

Proof

We use the fact that in a parallelogram the diagonals bisect each other, and that perpendicularity makes the half-sides equal.

1. Hypothesis. $ABCD$ a parallelogram with $AC\perp BD$. Let $O$ be the point of intersection of the diagonals.
2. Observe. The diagonals of a parallelogram bisect each other: $AO\cong OC$ and $BO\cong OD$.
3. Consider. In the right triangle $AOB$: $AB^2 = AO^2+OB^2$.
4. Similarly. $BC^2 = OB^2+OC^2 = OB^2+AO^2 = AB^2$.
5. Deduce. Hence $AB\cong BC$. By the parallelogram property ($AB\cong CD$, $BC\cong AD$), all four sides are congruent: $ABCD$ is a rhombus.

$\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Proof · Parallelogram · Rhombus
Skills: Proving · Synthetic geometry
People: Pythagoras