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

The correct answer is: Hence proved

Solution :-
Aim  :- Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.     where ,    and   are lengths of diagonals
Hint :- Draw the height from both A and B vertices .Applying pythagoras theorem we get the equation.Using these equations we prove the theorem.
Explanation(proof ) :-
As we know opposite sides of parallelogram are equal i.e AD = BC = L
AX and BY are perpendicular to line DC  and intersect at X and Y respectively.
Consider ΔADX and ΔBCY ,
AX = BY (distance between two parallel lines is always equal)
AD = BC (opposite sides of parallelogram are equal) (Hypotunes of triangles)
Both have right angles at X and Y .
By RHS congruence rule,

Let DX = CY = a and AC =  ; BD =  
Then ,XC = DC - DX =b-a ; DY = DC+CY = b+a
Applying pythagoras theorem in 

Applying pythagoras theorem in ΔBYD, we get

Adding equation 2 and 1 we get ,




Applying pythagoras theorem in ΔADX,

Substitute Eq4 in Eq3


Hence proved