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Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
To do:
We have to prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
Draw a circle with $Q$ as the centre.
The circle passes through $A, B$ and $O$ as shown in the figure.
This implies,
$QA = QB = QO$ (Radii of the circle)
We know that,
All the sides of a rhombus are equal.
This implies,
$AB = CD$
Multiplying by $\frac{1}{2}$ on both sides, we get,
$\frac{1}{2}AB = \frac{1}{2}CD$
$\Rightarrow AQ = DP$
$BQ = DP$
$Q$ is the midpoint of $AB$
This implies,
$AQ= BQ$.........(i)
Similarly,
$RA = SB$
Draw $PQ$ parallel to $AD$
$RA = QO$.........(ii)
From (i) and (ii), we get,
$QA = QB = QO$
Hence proved.
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