Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.


To do:

We have to prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.

Solution:

Let $ABCD$ be a rhombus.

Four circles are drawn on the sides $AB, BC, CD$ and $DA$ respectively.


$ABCD$ is a rhombus whose diagonals $AC$ and $BD$ intersect each other at $O$.

The diagonals of a rhombus bisect each other at right angles.

This implies,

$\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^o$

$\angle AOB = 90^o$ and a circle described on $AB$ as diameter will pass through $O$.

Similarly,

The circles on $BC, CD$ and $DA$ as diameters also pass through $O$.

Hence proved.

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Updated on: 10-Oct-2022

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