Prove that in any parallelogram, the sum of the square of all sides is equal to the sum of the squares of the diagonals.

In the parallelogram ABCD, AB = CD & AD = BC

Let DF and CE be two perpendiculars drawn on AB

In ΔAEC using Pythagoras theorem we can say

AC² = AE² + CE²

Since AE = AB + BE so we can say

⇒ AC² = (AB + BE)² + CE²

AC² = AB² + BE² + 2 × AB × BE + CE² …Equation (i)

In ΔDBF using Pythagoras theorem we can say

DB² = DF² + BF²

Since BF = AB - AF so we can say

⇒ DB² = (AB - AF)² + DF²

DB² = AB² + AF² - 2 × AB × AF + DF² …Equation (ii)

In ΔDAF & ΔCBE

DA = CB (Opposite sides of a parallelogram)

DF = CE (DCEF is a rectangle)

∠DFA = ∠ CEB (Perpendiculars)

So ΔDAF & ΔCBE are congruent by S.A.S. axiom of congruency

AF = BE (Corresponding Parts of Congruent Triangle)

Adding Equation (i) and (ii)

AC² + DB² = AB² + AF² - 2 × AB × AF + DF² + AB² + BE² + 2 × AB × BE + CE²

⇒ AC² + DB² = AB² + AF² + DF² + AB² + BE² + CE²

(Since AF = BE)

Since AB = CD (Opposite side of parallelogram)

⇒ AC² + DB² = AB² + AF² + DF² + CD² + BE² + CE²

Using Pythagoras theorem

⇒ AC² + DB² = AB² + AD² + CD² + BC²

Hence Proved