# 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 Δ AMD,

AD2 = DM2 + AM2 ...............eq(i)

In Δ BMD,

BD2 = DM2 + (AM + AB)2

Or,

(AM + AB)2 = AM2 + AB2 + 2 AM x AB

BD2 = DM2 + AM2 + AB2 + 2AM x AB ..................eq(ii)

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

BD2 = AD2 + AB2 + 2 x AM x AB .............eq(iii)

In Δ AND,

AD2 = AN2 + DN2     ......................eq(iv)

In Δ ANC,

AC2 = AN2 + (DC – DN)2

Or,

AC2 = AN2 + DN2 + DC2 – 2 x DC x DN ............eq(v)

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

AC2 = AD2 + DC2 – 2 x DC x DN ...............eq(vi)

We also have,

AM = DN and AB = CD

Substituting these values in (vi), we get

AC2 = AD2 + DC2 – 2 x AM x AB .................eq(vii)

Adding (iii) and (vii), we get

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

Or,

AC2 + BD2 = AB2 + BC2 + DC2 + AD2

Hence, proved.

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