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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.

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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.

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To prove: A circle drawn with Q as centre, will pass through A, B and O (i.e. QA = QB = QO)

Since all sides of a rhombus are equal,

AB = DC

Now, multiply (1/2) on both sides

(1/2)AB = (1/2)DC

AQ = DP

BQ = DP

Since Q is the midpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO we get,

QA = QB = QO (hence proved).

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