Question

prove that the sum of the square of the diagonals of a prarallelogram is equal to the sum of the squares of its side​

Answer 1

Step-by-step explanation:

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SOLUTION

Diagonals of a parallelogram bisect each other.

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.

Diagonals of a parallelogram bise

ct each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BD

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BD

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we get

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4⇒AB2+AD2+CB2+CD2=4AO2+4B

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4⇒AB2+AD2+CB2+CD2=4AO2+4B⇒AB2+AD2+CB2+CD2=(2AO)2+(

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4⇒AB2+AD2+CB2+CD2=4AO2+4B⇒AB2+AD2+CB2+CD2=(2AO)2+(⇒AB2+AD2+CB2+CD2=(AC)2+(B

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4⇒AB2+AD2+CB2+CD2=4AO2+4B⇒AB2+AD2+CB2+CD2=(2AO)2+(⇒AB2+AD2+CB2+CD2=(AC)2+(B⇒AB2+AD2+CB2+CD2=AC2+BD2

Diagonals of a parallelogram bisect each other.i.e. O is the mid point of AC and BD.In ∆ABD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem AB2+AD2=2AO2+2BO2(In ∆CBD, point O is the midpoint of side BD.BO=OD=12BDby Apollonius theorem CB2+CD2=2CO2+2BO2(Adding (1) and (2), we getAB2+AD2+CB2+CD2=2AO2+2BO⇒AB2+AD2+CB2+CD2=2AO2+4⇒AB2+AD2+CB2+CD2=4AO2+4B⇒AB2+AD2+CB2+CD2=(2AO)2+(⇒AB2+AD2+CB2+CD2=(AC)2+(B⇒AB2+AD2+CB2+CD2=AC2+BD2Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

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Answer 2

Answer;

In parallelogram ABCD,

AB=CD and BC=AD

Draw perpendiculars from C and D on AB as shown.

In right angled △AEC,

⇒ (AC)2=(AE)2+(CE)2

[ By Pythagoras theorem ]

⇒ (AC)2=(AB+BE)2 +(CE)2

⇒ (AC)2=(AB)2 +(BE)2+2×AB×BE+(CE)2

----- ( 1 )

From the figure CD=EF [ Since CDFE is a rectangle ]

But CD=AB

⇒ AB=CD=EF

Also CE=DF [ Distance between two parallel lines

⇒ △AFD≅△BEC [ RHS congruence rule ]

⇒ AF=BE [ CPCT ]

Considering right angled △DFB

⇒ (BD)2=(BF)2+(DF)2

[ By Pythagoras theorem ]

⇒ (BD)2=(EF−BE)2 +(CE)2

[ Since DF=CE ]

⇒ (BD)2=(AB−BE)2+(CE)2

[ Since EF=AB ]

⇒ (BD)2=(AB)2+(BE)2−2×AB×BE+(CE)2

----- ( 2 )

Adding ( 1 ) and ( 2 ), we get

(AC)2+(BD)2=(AB)2+(BE)2+2×AB×BE+(CE)2+(AB)2+(BE)2 −2×AB×BE+(CE) 2

⇒ (AC)2+(BD) 2=2(AB)2 +2(BE)2+2(CE)2

⇒ (AC)2+(BD)2=2(AB)2+2[(BE)2+(CE)2

] ---- ( 3 )

In right angled △BEC,

⇒ (BC)2=(BE)2 +(CE) 2

[ By Pythagoras theorem ]

Hence equation ( 3 ) becomes,

⇒ (AC)2+(BD)2=2(AB)2+2(BC)2

⇒ (AC)2+(BD)2=(AB)2 +(AB) 2+(BC)2+(BC)2

∴ (AC)2 +(BD)2=(AB)2+(BC)2+(CD)2+(AD)2

∴ The sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides.

hope it helps you