Topic:- Quadrilaterals
Class:- 9th
[tex] \sf \red{Question:}[/tex]
Show that the diagonals of a square are equal and bisect each other at right angles.
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Topic:- Quadrilaterals
Class:- 9th
[tex] \sf \red{Question:}[/tex]
Show that the diagonals of a square are equal and bisect each other at right angles.
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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.
Verified answer
[tex] \star \; {\underline{\boxed{\orange{\pmb{\frak{ \; Given \; :- }}}}}} [/tex]
[tex] \\ \\ [/tex]
[tex] \star \; {\underline{\boxed{\color{darkblue}{\pmb{\frak{ \; To \; Prove \; :- }}}}}} [/tex]
[tex] \\ \\ [/tex]
[tex] \star \; {\underline{\boxed{\pink{\pmb{\frak{ \; ProoF \; :- }}}}}} [/tex]
[tex] \underline{\pmb{\sf{ In \; ∆ADC \; \& \; BCD \; :- }}} [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { AD = BC } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { \angle ADC = \angle BCD } \qquad \; \; \bigg( Each \; {90}^{ \circ } \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { DC = DC } \qquad \; \; \bigg( Common \bigg) } \\ \\ [/tex]
[tex] \therefore \; [/tex] ∆ADC ≈ ∆BCD by SAS Congruence Criteria .
[tex] \\ [/tex]
[tex] \dag [/tex] So :
[tex] \qquad \; {\implies{\underline{\boxed{\purple{\pmb{\sf{ AC = BD }}}}} \sf \qquad ---- (i) }} [/tex]
[tex] \\ \\ [/tex]
[tex] \underline{\pmb{\sf{ Now, \; in \; ∆AOD \; \& \; ∆BOC \; :- }}} [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { AD = BC } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { \angle 1 = \angle 2 } \qquad \; \; \bigg( Alternate \; Interior \; Angles \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { \angle 3 = \angle 4 } \qquad \; \; \bigg( Alternate \; Interior \; Angles \bigg) } \\ \\ [/tex]
[tex] \therefore \; [/tex] ∆AOD ≈ ∆COB by ASA Congruence Criteria .
[tex] \\ [/tex]
[tex] \dag [/tex] So :
[tex] \qquad \; {\implies{\underline{\boxed{\red{\pmb{\sf{ AO = CO , BO = OD }}}}} \sf \qquad ---- (ii) }} [/tex]
[tex] \\ \\ [/tex]
[tex] \underline{\pmb{\sf{ Now, \; in \; ∆AOD \; \& \; ∆AOB \; :- }}} [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { AD = AB } \qquad \; \; \bigg( All \; Sides \; are \; Equal \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { DO = BO } \qquad \; \; \bigg( Proved \; Above \bigg) } \\ \\ [/tex]
[tex] \qquad \; {\dashrightarrow \; \sf { AO = AO } \qquad \; \; \bigg( Common \bigg) } \\ \\ [/tex]
[tex] \therefore \; [/tex] ∆AOD ≈ ∆CAOB by SSS Congruence Criteria .
[tex] \\ [/tex]
[tex] \dag [/tex] So :
[tex] \qquad \; {\implies{\underline{\boxed{\orange{\pmb{\sf{ \angle AOD = \angle AOB }}}}}}} [/tex]
[tex] \\ \\ [/tex]
[tex] \underline{\pmb{\sf{ But \; :- }}} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD + \angle AOB = {180}^{ \circ } } \qquad \bigg( Linear \; Pair \bigg) \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD + \angle AOD = {180}^{ \circ } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { 2 \angle AOD = {180}^{ \circ } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD = \dfrac{ {180}^{ \circ } }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { \angle AOD = \cancel\dfrac{ {180}^{ \circ } }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf { \angle AOD = {90}^{ \circ } }}}}} \; {\blue{\bigstar}} \\ \\ \\ \end{gathered} [/tex]
[tex] \\ [/tex]
[tex] \dag [/tex] So :
[tex] \qquad \; {\implies{\underline{\boxed{\pink{\pmb{\sf{ \angle AOD = {90}^{ \circ} }}}}}}} [/tex]
[tex] \\ \\ [/tex]
[tex] \therefore [/tex] From (i) , (ii) and (iii) we have Proved that the Diagonals are equal and bisect each other at right angles .
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]