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Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

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Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

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Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

DE2 + EA2 = DA2 ……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2

(DE2 + EA2) + AB2 + 2EA × AB = DB2

DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

AD2 = AF2 + FD2

Again, applying Pythagoras theorem in ∆AFC, we get,

AC2 = AF2 + FC2 = AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC × FD

= (AF2 + FD2) + DC2 − 2DC × FD. AC2

AC2= AD2 + DC2 − 2DC × FD ………………… (iii)

Since ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

In ∆DEA and ∆ADF,

∠DEA = ∠AFD (Each 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common Angles)

∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF ……………… (vi)

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

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

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

From equation (iv) and (vi),

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

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

This post was modified 3 years ago by Raavi Tiwari
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