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: ⇒〖d_1〗^2+〖d_2〗^2=2(l^2+b^2)

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,
ΔADX ≅ ΔBCY ; then DX = CY (by congruence)

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

Applying pythagoras theorem in ΔACX ,We


Applying pythagoras theorem in ΔBYD, we get
 — Eq2
Adding equation 2 and 1 we get ,
 — Eq3

Applying pythagoras theorem in ΔADX,
—- Eq4
Substitute Eq4 in Eq3
We get