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$\sf \red{Question:}$
☆ Show that if the diagonals of the quadrilateral are equal and bisect each other at right angles, then it is a square.

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Answer 1

$\star \; {\underline{\boxed{\pmb{\orange{\frak{ \; Given \; :- }}}}}}$

• ABCD is a Quadrilateral
• AC = BD
• OA = OC and OB = OD
• ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

$\\ {\underline{\rule{200pt}{3pt}}}$

$\star \; {\underline{\boxed{\pmb{\color{darkblue}{\frak{ \; To \; Prove \; :- }}}}}}$

• ABCD is a Square

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$\star \; {\underline{\boxed{\pmb{\red{\frak{ \; ProoF \; :- }}}}}}$

Here :-

${\underline{\textbf{\textsf{ In \; ∆AOB \; and \; ∆COB \; : }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { OA = OC \qquad \; \bigg( Given \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠AOB = ∠COB \qquad \; \bigg( Each \; 90° \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { OB = OB \qquad \; \bigg( Common \bigg) } \\ \end{gathered}$

$\therefore \;$ ∆AOB ≈ ∆COB by SAS Congruence Criteria .

Hence :-

$\begin{gathered} \implies \; \; {\underline{\boxed{\pmb{\sf{ AB = CB }}}}} \; {\purple{\bigstar}} \end{gathered}$

$\\ \qquad{\rule{150pt}{1pt}}$

Similarly :-

${\underline{\textbf{\textsf{ In \; ∆AOB \; and \; ∆DOA \; : }}}}$

$\therefore \;$ ∆AOB ≈ ∆DOA by SAS Congruence Criteria .

Hence :-

$\begin{gathered} \implies \; \; {\underline{\boxed{\pmb{\sf{ AB = AD }}}}} \; {\green{\bigstar}} \end{gathered}$

$\\ \qquad{\rule{150pt}{1pt}}$

Similarly :-

${\underline{\textbf{\textsf{ In \; ∆BOC \; and \; ∆COD \; : }}}}$

$\therefore \;$ ∆BOC ≈ ∆COD by SAS Congruence Criteria .

Hence :-

$\begin{gathered} \implies \; \; {\underline{\boxed{\pmb{\sf{ CB = DC }}}}} \; {\orange{\bigstar}} \end{gathered}$

$\therefore \;$ We have proved that ABCD is a Parallelogram .

$\\ \qquad{\rule{150pt}{1pt}}$

Here :-

${\underline{\textbf{\textsf{ In \; ∆ABC \; and \; ∆DCB \; : }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { AC = BD \qquad \; \bigg( Given \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { AB = DC \qquad \; \bigg( Opp. \; Sides \; are \; Eql \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { BC = BC \qquad \; \bigg( Common \bigg) } \\ \end{gathered}$

$\therefore \;$ ∆ABC ≈ ∆DCB by SSS Congruence Criteria .

Hence :-

$\begin{gathered} \implies \; \; {\underline{\boxed{\pmb{\sf{ ∠ABC = ∠DCB }}}}} \; {\blue{\bigstar}} \end{gathered}$

Now :-

${\underline{\textbf{\textsf{ AB \; || \; CD \; and \; BC \; is \; the \; Transversal \; : }}}}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠ABC + ∠DCB = 180° \qquad \; \bigg( Alternate \; Interior \; Angles \bigg) } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠ABC + ∠DCB = 180° } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠ABC + ∠ABC = 180° } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { 2∠ABC = 180° } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠ABC = \dfrac{180}{2} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { ∠ABC = \cancel\dfrac{180}{2} } \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf{ ∠ABC = 90° }}}}} \; {\red{\bigstar}} \end{gathered}$

$\\ \qquad{\rule{150pt}{1pt}}$

$\therefore \;$ ABCD is a Square with one angle 90° .

$\\ {\underline{\rule{300pt}{9pt}}}$

Answer 2

$\huge\star{\blue{\underline{\underline{\tt{Explanation:}}}}}$

$\rule{210pt}{0.5pt}$

Given that,

• Let ABCD be a quadrilateral
• It's iagonals AC and BD bisect each other at right angle at O.

To prove that,

• The Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

$\leadsto$ AO = CO (Diagonals bisect each other)

$\leadsto$ ∠AOB = ∠COD (Vertically opposite)

$\leadsto$ OB = OD (Diagonals bisect each other)

$,\leadsto$ ΔAOB ≅ ΔCOD [SAS congruency]

Thus,

$\leadsto$ AB = CD [CPCT] — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

$\leadsto$ In ΔAOD and ΔCOD,

$\leadsto$ AO = CO (Diagonals bisect each other)

$\leadsto$ ∠AOD = ∠COD (Vertically opposite)

$\leadsto$ OD = OD (Common)

$\leadsto$ ΔAOD ≅ ΔCOD [SAS congruency]

Thus,

AD = CD [CPCT] ____ (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB ____ (ii)

also, ∠ADC = ∠BCD [CPCT]

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° ____ (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

$\large\star{\tt{\red{\underline{Hence\:Proved!}}}}$

$\begin{gathered} \\ {\underline{\rule{300pt}{9pt}}} \end{gathered}$