Question

prove that the sum of square of the diagonals of parallelogram is equal to the sum of square of its sides​

Answer 1

In parallelogram ABCD, AB = CD, BC = AD Draw perpendiculars from C and D on AB as shown. In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem] ⇒ AC2 = (AB + BE)2 + CE2 ⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2 → (1) From the figure CD = EF (Since CDFE is a rectangle) But CD= AB ⇒ AB = CD = EF Also CE = DF (Distance between two parallel lines) ΔAFD ≅ ΔBEC (RHS congruence rule) ⇒ AF = BE Consider right angled ΔDFB BD2 = BF2 + DF2 [By Pythagoras theorem] = (EF – BE)2 + CE2 [Since DF = CE] = (AB – BE)2 + CE2 [Since EF = AB] ⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2 → (2) Add (1) and (2), we get AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2) = 2AB2 + 2BE2 + 2CE2 AC2 + BD2 = 2AB2 + 2(BE2 + CE2) → (3) From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem] Hence equation (3) becomes, AC2 + BD2 = 2AB2 + 2BC2 = AB2 + AB2 + BC2 + BC2 = AB2 + CD2 + BC2 + AD2 ∴ AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Hope that this answer will help you ✌️

Answer 2

Given:

• A parallelogram PQRS.
• Diagonals are QS and PR
• Sides are PQ , QR , RS and SP.

To Prove:

• Sum of square of the diagonal of parallogram is equal to the sum of square of its sides. Means

• PR² + QS² = PQ² + QR² + RS² + SP²

Proof: In any ∆ABC if AD is a median of then

• AB² + AC² = 2 (AD² + BD²) i.e
• AB² + AC² = 2AD² + 1/2 BC².........(1)

Now let O be the point of intersection of the diagonal PR and QS.

• We know that the diagonals of a Parallelogram bisect each other •

∴ O is the midpoint of PR as well as QS.

Applying result (1) to ∆PQR and ∆RSP, we get

$\small\implies{\sf }$ PQ² + QR² = 2OQ² + 1/2PR²........(2)

$\small\implies{\sf }$ RS² + SP² = 2OS² + 1/2PR²........(3)

• Add the equations 2 and 3 •

$\small\implies{\sf }$ PQ² + QR² + RS² + SP² = 2(1/2QS)² + 2(1/2QS)² + PR²

$\small\implies{\sf }$ PQ² + QR² + RS² + SP² = PR² + QS²

Hence, Proved!