Question

Topic:- Quadrilaterals
Class:- 9th

$\sf \red{Question:}$
Show that the diagonals of a square are equal and bisect each other at right angles.
​

Answer 1

Answer:

Given that ABCD is a square

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180° (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.

Answer 2

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

• ABCD is a Square .

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

• AC = BD
• AO = CO , BO = DO
• ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

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

$\underline{\pmb{\sf{ In \; ∆ADC \; \& \; BCD \; :- }}}$

$\qquad \; {\dashrightarrow \; \sf { AD = BC } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) }$

$\qquad \; {\dashrightarrow \; \sf { \angle ADC = \angle BCD } \qquad \; \; \bigg( Each \; {90}^{ \circ } \bigg) }$

$\qquad \; {\dashrightarrow \; \sf { DC = DC } \qquad \; \; \bigg( Common \bigg) }$

$\therefore \;$ ∆ADC ≈ ∆BCD by SAS Congruence Criteria .

$\dag$ So :

$\qquad \; {\implies{\underline{\boxed{\purple{\pmb{\sf{ AC = BD }}}}} \sf \qquad ---- (i) }}$

$\underline{\pmb{\sf{ Now, \; in \; ∆AOD \; \& \; ∆BOC \; :- }}}$

$\qquad \; {\dashrightarrow \; \sf { AD = BC } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) }$

• AD || BC and AC is the Transversal .So :

$\qquad \; {\dashrightarrow \; \sf { \angle 1 = \angle 2 } \qquad \; \; \bigg( Alternate \; Interior \; Angles \bigg) }$

• AD || BC and BD is the Transversal .So :

$\qquad \; {\dashrightarrow \; \sf { \angle 3 = \angle 4 } \qquad \; \; \bigg( Alternate \; Interior \; Angles \bigg) }$

$\therefore \;$ ∆AOD ≈ ∆COB by ASA Congruence Criteria .

$\dag$ So :

$\qquad \; {\implies{\underline{\boxed{\red{\pmb{\sf{ AO = CO , BO = OD }}}}} \sf \qquad ---- (ii) }}$

$\underline{\pmb{\sf{ Now, \; in \; ∆AOD \; \& \; ∆AOB \; :- }}}$

$\qquad \; {\dashrightarrow \; \sf { AD = AB } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) }$

$\qquad \; {\dashrightarrow \; \sf { DO = BO } \qquad \; \; \bigg( Proved \; Above \bigg) }$

$\qquad \; {\dashrightarrow \; \sf { AO = AO } \qquad \; \; \bigg( Common \bigg) }$

$\therefore \;$ ∆AOD ≈ ∆CAOB by SSS Congruence Criteria .

$\dag$ So :

$\qquad \; {\implies{\underline{\boxed{\orange{\pmb{\sf{ \angle AOD = \angle AOB }}}}}}}$

$\underline{\pmb{\sf{ But \; :- }}}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD + \angle AOB = {180}^{ \circ } } \qquad \bigg( Linear \; Pair \bigg) \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD + \angle AOD = {180}^{ \circ } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { 2 \angle AOD = {180}^{ \circ } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD = \dfrac{ {180}^{ \circ } }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD = \cancel\dfrac{ {180}^{ \circ } }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf { \angle AOD = {90}^{ \circ } }}}}} \; {\blue{\bigstar}} \\ \\ \\ \end{gathered}$

$\dag$ So :

$\qquad \; {\implies{\underline{\boxed{\pink{\pmb{\sf{ \angle AOD = {90}^{ \circ} }}}}}}}$

$\therefore$ From (i) , (ii) and (iii) we have Proved that the Diagonals are equal and bisect each other at right angles .

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