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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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