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

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$\bf \red{Answer}$

$\small \tt{Given : ABCD \: is \: a \: rhombus . \: A \: circle \: is \: drawn \: with \: AB \: as \: diameter.}$

$\tt \small \: {To \: prove \: : \: Circle \: passes \: through \: O, \: Point \: of \: intersection } \\ \tt \small{of \: diagonals \: AC \: and \: BD.}$

$\tt \small{Proof : \: We \: know \: that , \: the \: diagonals \: of \: rhombus \: bisect \:} \\ \tt \small{ each \: other \: at \: 90 \degree}$

$\tt \small{ \therefore \angle \: AOB = 90 \degree}$

$\tt \small {But \: angle \: in \: semi-circle \: is \: 90 \degree}$

$\tt \small{Thus, Circle \: passes \: through \: O.}$

$\tt \: \small \: {Therefore , \: any \: circle \: drawn \: with \: any \: side \: of \: a \: rhombus} \\ \tt \small \: {as \: diameter \: , passes \: through \: the \: point \: of \: } \\ \tt \small{ intersection \: of \: its \: diagonals \: } .$

$\tt \bf {Hence , Proved \: ✔}$

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Answer 2

answer is in attachment