Question

Prove that the sum of square of the diagonals of a parallelogram is equal to the sum of square of its sides?


Plz answer

Answer 1

Let ABCD be aparallelogram.

Let us drawperpendicular DE on extended side AB, and AF on side DC.

Applying Pythagorastheorem in ΔDEA,we obtain

DE2+ EA2= DA2 …(i)

Applying Pythagorastheorem in ΔDEB,we obtain

DE2+ EB2= DB2

DE2+ (EA + AB)2= DB2

(DE2+ EA2)+ AB2+ 2EA × AB = DB2

DA2+ AB2+ 2EA × AB = DB2 …(ii)

Applying Pythagorastheorem in ΔADF, we obtain

AD2 = AF2+ FD2

Applying Pythagorastheorem in ΔAFC, we obtain

AC2 = AF2+ FC2

=AF2 + (DC − FD)2

=AF2 + DC2 + FD2 − 2DC ×FD

=(AF2 + FD2) + DC2 − 2DC ×FD

AC2 = AD2+ DC2 − 2DC × FD … (iii)

Since ABCD is aparallelogram,

AB = CD …(iv)

And, BC = AD …(v)

In ΔDEAand ΔADF,

∠DEA =∠AFD (Both 90°)

∠EAD = ∠ADF (EA|| DF)

AD = AD (Common)

∴ ΔEAD ΔFDA (AAS congruencecriterion)

⇒ EA = DF …(vi)

Adding equations (i)and (iii), we obtain

DA2+ AB2+ 2EA × AB + AD2+ DC2− 2DC × FD = DB2+ AC2

DA2+ AB2+ AD2+ DC2+ 2EA × AB − 2DC × FD = DB2+ AC2

BC2 + AB2+ AD2 + DC2 + 2EA × AB − 2AB ×EA = DB2 + AC2

[Using equations (iv)and (vi)]

AB2+ BC2+ CD2+ DA2= AC2+ BD2

Answer 2

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.