Question

2) Prove that sum of the squares of the diagonals of parallelogram is equal to
the sum of the squares of it's sides.​

Answer 1

Answer:

O is the mid point of AC and BD. In ∆ABD, point O is the midpoint of side BD. In ∆CBD, point O is the midpoint of side BD. Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer 2

◘ Statement:-

• The sum of the squares of the diagonals in a Parallelogram is always equal to the sum of squares of it's sides.

◘ Given:-

• PQRS is Parallelogram.
• PQ,QR,RS & SP are sides of the parallelogram.
• PR & QS are diagonals.
• O is the midpoint of PR & QS, such that OP = OR & OQ = OS.

◘ Required To Prove:-

• (PQ)²+(QR)²+(RS)²+(SP)² = (PR)²+(QS)²

◘ Proof:-

In ∆PQR,

QO is the Median of the ∆le

WKT

☞ Diagonals of a Parallelogram bisect each other.

By Apollonius Theorem,

• PQ²+QR² = 2OQ²+2OP² ...[1]

In ∆PSR

SO is the median of the ∆le

WKT

☞ Diagonals of a Parallelogram bisect each other.

By Apollonius Theorem

• PS²+SR² = 2OS²+2OR² ...[2]

By Adding (1) & (2)

➝ PQ²+QR²+PS²+SR² = 2OQ²+2OP²+2OS²+2OR²

➝ PQ²+QR²+RS²+PS² = 2OP²+2OQ²+2OQ²+2OP²

➝ PQ²+QR²+RS²+PS² = 4OQ²+4OP²

As, The diagonals of a Parallelogram bisect each other.

• OP = PR/2 & OQ = QS/2
• OP = 1/2(PR) & OQ = 1/2(QS)

➝ PQ²+QR²+RS²+PS² = 4[1/2(PR)]²+4[1/2(QS)]²

➝ PQ²+QR²+RS²+PS² = [4(1/2)²(PR)²]+[4(1/2)²(QS)²]

➝ PQ²+QR²+RS²+PS² = 4×1/4 [(PR)²]+4×1/4 [(QS)²]

➝ PQ²+QR²+RS²+PS² = 4/4 (PR)²+ 4/4 (QS)²

➝ PQ²+QR²+RS²+PS² = 1(PR)²+1(QS)²

➝ PQ²+QR²+RS²+PS² = (PR)²+(QS)²

• PQ²+QR²+RS²+PS² = PR²+QS²

Hence Proved.

ꈍꈍꈍꈍꈍꈍꈍꈍꈍꈍ

☞ In a Parallelogram opposite sides are parallel and equal.

☞ In a Parallelogram opposite angles are equal and adjacent angles are supplementary.

☞ Apollonius Theorem states that the sum of squares of any two sides of any triangle equal twice the square on the half of third side, together twice the square on the median bisecting the third side.

@MrSovereign ツ

Hope This Helps!!