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