Question

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

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Solution

## $\mathrm{Given}:\mathrm{A}\mathrm{rhombus}\mathrm{ABCD}\mathrm{and}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{diameter}\mathrm{AB}.\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{prove}:\mathrm{Point}\mathrm{O}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circle}\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}\mathrm{As},\mathrm{angle}\mathrm{formed}\mathrm{in}\mathrm{the}\mathrm{semi}-\mathrm{circle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}.\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOB}=90°\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{diagonals}\mathrm{intersect}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{a}\mathrm{right}\mathrm{angle}.\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOB}=\angle \mathrm{BOC}=\angle \mathrm{COD}=\angle \mathrm{DOA}=90°\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{conlcuded}\mathrm{that}\mathrm{point}\mathrm{O}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circle}.$ Hence, the circle drawn with any side of rhombus as diameter passes through the point of intersection of its diagonals.

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