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

In parallelogram ABCD, Altitudes AF and DE is drawn on DC and produced BA.
In ΔDEA, by pythagoras theorem, DE² + EA² = DA² = …(1)
In ΔDEB. by pythagoras theorem, DE² + EB² = DB²
⇒ DE² + (EA + AB)² = DB²
⇒ (DE² + EA²) + AB² + 2EA × AB = DB²
⇒ DA² + AB² + 2EA × AB = DB² …(ii)
In ΔADF, by Pythagoras theorem, AD² = AF² + FD²
In ΔAFC, by pythagoras theorem
AC² = AF² + FC² = AF² + (DC – FD)² = AF² + DC² + FD² – 2DC × FD
= (AF² + FD²) + DC² – 2DC × FD
⇒ AC² = AD² + DC² – 2DC × FD …(iii)
ABCD is a parallelogram.
Therefore
AB = CD …(iv)
and, BC = AD …(v)
In ΔDEA and ΔADF,
∠DEA = ∠AFD [Each 90]
∠EAD = ∠ADF [EA II DF]
AD = AD [Common]
∴ΔEAD ≅ ΔFDA [AAS congruency rule]
⇒ EA = DF …(vi)
Adding equations (ii) and (iii), we have
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 [From the equation (iv) and (vi)]
⇒ AB² + BC² + CD² + DA² = AC² + BD²