Diagonals of Rhombus Bisect Each Other at Right Angles
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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 $2$