Question

Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point


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Answer 1

Answer:

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.

Recall that, diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Consider the circle with AB as diameter passes through O. [Since angle in a semi-circle is a right angle]

Similarly we can prove for other sides BC, CD and DA also.

Therefore, the circles with four sides of a rhombus as diameter, pass through the point of intersection of its diagonals

Answer 2

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$\bf \red{Answer}$

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

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

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

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

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

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

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

$\tt \bf {Hence , Proved \: ✔}$

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