Question

prove that the circles described on the four sides of a rhombus as diameters, pass
the point of intersection of its diagonals.
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Answer 1

$\huge\mathbb{SOLUTION:-}$

• Given :-

• A circle which is made of 4 sides of rhombus

• We have to prove that the Diameter of rhombus passes through the point of intersection of its diagonal

Now

$\bf\underline{\red{According\:To\: Question:-}}$

Let PQRS be a rhombus and O be the point of intersection of diagonal

We know that the diagonal of rhombus intersect at 90° .

So :-

$\sf \angle PQR=\angle QRS=\angle RSP =\angle SPQ=90{}^{\circ}$

So all the angles of rhombus=90°

$\sf\implies \angle POQ=90{}^{\circ}$

O is the point of intersection

And OQ is the radius of circle .. ... (i)

SO is also the radius of circle. .. (ii)

Now add eq. (i) and (ii)

SO+OQ

=SQ (which is the diameter of circle and diagonal of rhombus)

So the diameter of circle passes through the diagonal of rhombus.. (Hence Proved !! ! )

Answer 2

Step-by-step explanation:

Sol: Consider ABCD be a rhombus whose diagonals AC and BD intersect at O.

Since diagonals of a rhombus intersect each other at right angle.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, AB, BC, CD and DA as diameters of circles passes through O. Angle in a semi-circle is 90°.

Hence the circles described on the four sides of a rhombus as diameter, passes through the point of intersection of its diagonals.