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

 ABCD is a parallelogram in which AB = CD and AD = BC

Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extend up to M

In triangle AMD,

AD² = DM² + AM²     (i)

In triangle BMD,

BD² = DM² + (AM + AB)²

Or,

BD² = DM² + AM² + AB² + 2AM * AB     (ii)

Substituting the value of AM² from (i) in (ii), we get

BD² = AD² + AB² + 2 * AM * AB    (iii)

In triangle AND,

AD² = AN² + DN²

In triangle ANC,

AC² = AN² + (DC – DN)²

Or,

AC² = AN² + DN² + DC² – 2 * DC * DN     (v)

Substituting the value of AD² from (iv) in (v), we get

AC² = AD² + DC² – 2 * DC * DN     (vi)

We also have,

AM = DN and AB = CD

Substituting these values in (vi), we get

AC² = AD² + DC² – 2 * AM *- AB     (vii)

Adding (iii) and (vii), we get

AC² + BD² = AD² + AB² + 2 * AM * AB + AD² + DC² – 2 * AM * AB

Or,

AC² + BD² = AB² + BC² + DC² + AD²

Hence, proved