Question

Prove that the figure obtained by joining the mid points of the adjacent sides of a rectangle is a rhombus

Answer 1

$\boxed { \huge \mathfrak\red{{Answer }}}$

Given :

A rectangle in which P , Q , R , S are the mid points of S

AB , BC , CD , DA

Follow according to the attached figure

To prove :

PQ parallel to RS

PQ = RS = QR = SP

Construction :

Draw the diagonal of the rectangle which are equal to each other

Proof :

In a rectangle ABCD

$\implies$ BD = AC

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In a triangle ABC

P is the mid point of AB

Q is the mid point of BC

Then PQ // AC , PQ = $\dfrac{1}{2}$AC ( By the mid point of converse theory )

$\implies$PQ =$\dfrac{1}{2}$ BD , PQ // AC --- ( i )

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In triangle ADC

S is the mid point of AD

R is the mid point of DC

then SR // AC

SR = $\dfrac{1}{2}$AC ( By the mid point of converse theory )

SR = $\dfrac{1}{2}$ BD , SR // AC -- ( ii )

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$\implies$In triangle ADB

S is the mid point of AD

P is the mid point of AB

then PS // DB

PS = $\dfrac{1}{2}$ DB ( By the mid point of converse theory)

PS // DB , PS = $\dfrac{1}{2}$ DB -- ( iii )

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$\implies$In triangle BDC

R is the mid point of DC

Q is the mid point of BC

then QR // BD

QR = $\dfrac{1}{2}$BD ( By the mid point of converse theory )

$\implies$QR // BD , QR = $\dfrac{1}{2}$BD -- ( iv )

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From i , ii , iii , iv

$\implies$PQ // AC , PQ = $\dfrac{1}{2}$ BD

$\implies$SR // AC , SR = $\dfrac{1}{2}$ BD

$\implies$PS // BD , PS = $\dfrac{1}{2}$BD

$\implies$QR // BD , QR = $\dfrac{1}{2}$ BD

PS // SR , PS // QR , PS = QR = PS = QR

According to the rule In a Rhombus, opposite sides are always parallel and all the adjacent sides are equal ! so here it happens

$\implies$So it is RHOMBUS !

HENCE PROVED