#### Answer

We consider a rectangle. where the sides are a, b, c, and d and where the diagonals are A and B. Thus, we obtain:
$A^2 = a^2 + b^2 $
And:
$B^2 = c^2 + d^2 $
Adding these gives:
$ A^2 + B^2 = a^2 + b^2 +c^2 + d^2$
Thus, the proof is completed.

#### Work Step by Step

In a $ parallelogram $ ABCD, we have to prove that diagonals $DB^{2} + AC^{2} = AB^{2}+BC^{2}+DC^{2} + AD^{2}$.
1-Drawing two altitudes AT where T is a point on segment DC , and BV where V is an exterior point on the opposite line.
2-proving triangle DAT = triangle BCV by LH theorem, therefore AT= BV, DT= CV.
3- let AB= DC= b , AD= BC= a, and let TC = c.
4- applying Pythagoras theorem on triangle BVD, $BD^{2} = DV^{2} + BV^{2}$
And applying it on triangle ATC
$AC^{2} = AT^{2} + TC^{2}$
5- writing segments in term of a,b and c
Segment DV= AB + DC -TC = 2b-c. ( segment subtraction method )
Segment $AT^{2}=BV^{2}= a^{2} - ( b-c )^2 $
6- by substituting in the Pythagoras theorems above such that
$ BD^{2} = (a^{2} - ( b-c)^2 ) + ( 2b-c)^2 $
And
$ AC^{2}= (a^2 -(b-c)^2) + c^{2} $
So
$ BD^2 + AC^2 = 2 ( a^2 -(b-c)^2) + c^2 + ( 2b-c)^2=.....= 2( a^2 + b^2) = a^2+b^2+a^2+b^2 $
Since a= AD= BC and b= AB= DC.
So we prove that $ AB^2+ AC^2= AB^2+BC^2+DC^2+AD^2 $
$\square$