Question

Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.​

Answer 1

Answer:

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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)

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Since all sides of a rhombus are equal,:-

• AB = DC

Now,

Multiply (½) on both sides

• (½)AB = (½)DC
• So, AQ = DP

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⇒ 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,

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• as AQ = BQ and RA = QO we have,
• QA = QB = QO (hence proved).

Answer 2

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To prove:

A circle drawn with Q as center , will pass through AB and O

We know that,

All sides of the Rhombus are equal

:AB = BC = CD = DA

=AB = CD

multiple 1/2 on both side

• (1/2)AB = (1/2)CD

• so AQ=CP

That

BQ = DP

since Q is the mid point of AB

AQ = BQ

similarly,

RA = SB

Again,

as PQ is drawn parallel to AD

RA = QO

Now,

As

AQ = BQ AND RA = QO

QA = QB = QO

Hence proved

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