Question

Prove that the diagonals of a rhombus bisect each other at right angles.

Answer 1

Rhombus is a parallelogram.

Consider:

$$\begin{gathered} ∆AOB\mathrm{and}∆COD \\ \angle OAB=\angle OCD(\mathrm{alternate}\mathrm{angle}) \\ \angle ODC=\angle OBA(\mathrm{alternate}\mathrm{angle}) \\ \angle DOC=\angle AOB(\mathrm{vertically}\mathrm{opposite}\mathrm{angles}) \\ ∆AOB\cong COB \\ \therefore AO=CO \\ OB=OD \end{gathered}$$

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider $∆COD\hspace{0.17em}and∆COB$:
$$\begin{gathered} CD=CB\hspace{0.17em}(\mathrm{all}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{are}\mathrm{equal}) \\ CO=CO(\mathrm{common}\mathrm{side}) \\ OD=OB\hspace{0.17em}(p\mathrm{oint}O\mathrm{bisects}BD) \end{gathered}$$

∴ $∆COD\cong ∆COB$
∴ $\angle COD=\angle COB$ (corresponding parts of congruent triangles)

Further, $\angle COD+\angle COB=180°(l\mathrm{inear}\mathrm{pair})$

∴ $\angle COD=\angle COB=90°$

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.