Question

A rhombus is a quadrilateral that has four congruent sides. How would you prove that the diagonals of a rhombus intersect at a point that is the midpoint of each diagonal?

Answer 1

See the proof below

Explanation:

Assume a rhombus `ABCD` with diagonals `AC` and `BD` intersecting at point `O`.

Triangles `Delta ABD` and `Delta CBD` are congruent by three sides. Therefore, angles `/_ABD` and `/_CBD` are congruent.

Consider now triangles `Delta ABO` and `Delta CBO`. They are also congruent since `AB~=CB`, `BO` is common and angles `/_ABD` and `/_CBD` are congruent (side-angle-side).
Therefore, `AO~=CO`.

Similarly `BO~=DO` if you compare triangles `Delta BAC` and `Delta DAC`, then angles `/_BAC` and `/_DAC` and, finally, triangles `Delta BAO` and `Delta DAO`.