Maths-

General

Easy

Question

# 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

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