Question

Show that if the diagnals of a quardrilateral bisect each other at right angles, then it is a rhombus.
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Answer 1

Answer:

annyeonghaseyo unnie

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at the right angle. So, we have, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.

might help you

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Answer 2

Given :

• i) AO = OC
• ii) OB = OD
• iii) ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

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To Prove :

• i) ABCD is a Rhombus .

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Proof :

In ∆AOB and ∆BOC :

${\longmapsto{\qquad{ \: \: {\sf{ AO = CO }}}}}{\: \: \: \: \: \: \: \: \: \: { \lgroup {\red{\sf{Given}}}} \rgroup }$

${\longmapsto{\qquad{ \: \: {\sf{ OB = OB }}}}}{\: \: \: \: \: \: \: \: \: \: \: { \lgroup {\red{\sf{Common}}}} \rgroup }$

${\longmapsto{\qquad{ \: \: {\sf{ ∠AOB = ∠BOC }}}}}{\: \: \: { \lgroup {\red{\sf{Each \: 90°}}}} \rgroup }$

Hence,

∆AOB ≈ ∆BOC by SAS Congruence criteria .

So,

$\large{\underline{\boxed{\sf{ AB = BC }}}} \: \: \: \: { \lgroup {\orange{\sf{CPCT}}} \rgroup } --- {\purple{\sf{ (i) }}}$

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Similarly ,

∆COD ≈ ∆AOD

So,

$\large{\underline{\boxed{\sf{ CD = AD }}}} \: \: \: \: { \lgroup {\orange{\sf{CPCT}}} \rgroup } --- {\purple{\sf{ (ii) }}}$

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Similarly ,

∆BOC ≈ ∆COD

So,

$\large{\underline{\boxed{\sf{ BC = CD }}}} \: \: \: \: { \lgroup {\orange{\sf{CPCT}}} \rgroup } --- {\purple{\sf{ (iii) }}}$

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Now ,

From (i) ,(ii) and (iii) :

${\dashrightarrow{\sf{ AB = BC = CD = AD }}}{\qquad{\: \: {\lgroup {\orange{\sf{From \: (i) \;,\:(ii) \: and \: (iii) }}} \rgroup }}}$

Proved .

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Hence ,

ABCD is a Rhombus .

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