Question

prove that the sums of the squares of the daigonal of a parallelogran is equal to the sum of the squares of its sides
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Answer 1

Answer:

It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. ... If the parallelogram is a rectangle, the two diagonals are of equal lengths (AC) = (BD) so, and the statement reduces to the Pythagorean theorem.

Step-by-step explanation:

In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. Using the notation in the diagram on the right, the sides are (AB), (BC), (CD), (DA). But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, or (AB) = (CD) and (BC) = (DA), the law can be stated as,

{\displaystyle 2(AB)^{2}+2(BC)^{2}=(AC)^{2}+(BD)^{2}\,} 2(AB)^2+2(BC)^2=(AC)^2+(BD)^2\,

If the parallelogram is a rectangle, the two diagonals are of equal lengths (AC) = (BD) so,

{\displaystyle 2(AB)^{2}+2(BC)^{2}=2(AC)^{2}\,} 2(AB)^2+2(BC)^2=2(AC)^2\,

and the statement reduces to the Pythagorean theorem. For the general quadrilateral with four sides not necessarily equal,

{\displaystyle (AB)^{2}+(BC)^{2}+(CD)^{2}+(DA)^{2}=(AC)^{2}+(BD)^{2}+4x^{2},\,} {\displaystyle (AB)^{2}+(BC)^{2}+(CD)^{2}+(DA)^{2}=(AC)^{2}+(BD)^{2}+4x^{2},\,}

where x is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that, for a parallelogram, x = 0, and the general formula simplifies to the parallelogram law.