answer fast.
.
.
..
...
.
.
.
Answers
To prove this statement, first we have to recall the law of cosines.
It states that, in ∆ABC of sides a, b, c, if two sides and the angle between them are known, then the third side is given by the following possible equations.
a² = b² + c² - 2bc cos A
b² = c² + a² - 2ac cos B
c² = a² + b² - 2ab cos C
where the sides a, b, c each should be opposite to the angles A, B and C respectively.
From the fig., since ABCD is a parallelogram, we have,
AB = CD → (1)
AD = BC → (2)
And it is true in every parallelogram that the angles at two ends of any side of the parallelogram are always supplementary. Thus from the parallelogram ABCD,
⟨ABC + ⟨BCD = 180°
⟨BCD = 180° - ⟨ABC
Taking cosines on both sides,
cos ⟨BCD = cos (180° - ⟨ABC)
But cos (180° - x) = - cos x. Thus,
cos ⟨BCD = - cos ⟨ABC → (3)
By law of cosines, from ∆ABC,
(AC)² = (AB)² + (BC)² - 2(AB)(BC) cos ⟨ABC → (4)
And from ∆BCD,
(BD)² = (BC)² + (CD)² - 2(BC)(CD) cos ⟨BCD
But from (1), (2) and (3),
(BD)² = (AD)² + (CD)² + 2(AB)(BC) cos ⟨ABC → (5)
Now, adding (4) and (5), we get,
(AC)² + (BD)² = (AB)² + (BC)² + (CD)² + (AD)²
This equation says that the sum of squares of sides of a parallelogram is equal to that of the squares of diagonals of the parallelogram.
Hence Proved!
#answerwithquality
#BAL
