Question

Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{A}\mathrm{rhombus}\mathrm{ABCD}\mathrm{and}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{diameter}\mathrm{AB}. \\ \mathrm{To}\mathrm{prove}:\mathrm{Point}\mathrm{O}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circle} \\ \mathrm{Proof}: \\ \mathrm{As},\mathrm{angle}\mathrm{formed}\mathrm{in}\mathrm{the}\mathrm{semi}-\mathrm{circle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}. \\ \Rightarrow \angle \mathrm{AOB}=90° \\ \mathrm{Also},\mathrm{diagonals}\mathrm{intersect}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{a}\mathrm{right}\mathrm{angle}. \\ \Rightarrow \angle \mathrm{AOB}=\angle \mathrm{BOC}=\angle \mathrm{COD}=\angle \mathrm{DOA}=90° \\ \mathrm{So},\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{conlcuded}\mathrm{that}\mathrm{point}\mathrm{O}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circle}. \end{gathered}$$

Hence, the circle drawn with any side of rhombus as diameter passes through the point of intersection of its diagonals.