Diagonals of Rhombus Bisect Each Other at Right Angles

Theorem

Let $ABCD$ be a rhombus.

The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles.

Proof

By the definition of a rhombus, $AB = AD = BC = DC$.

Without loss of generality, consider the diagonal $BD$.

Thus:

$\triangle ABD$ is an isosceles triangle whose apex is $A$ and whose base is $BD$.

By Diagonals of Rhombus Bisect Angles, $AC$ bisects $\angle BAD$.

From Bisector of Apex of Isosceles Triangle also Bisects Base, $AC$ bisects $BD$.

From Bisector of Apex of Isosceles Triangle is Perpendicular to Base, $AC$ bisects $BD$ at right angles.

Hence the result.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.22$: Corollary $2$