prove that the sum of the square of the side of a rhombus is equal to the sum of the square of its diagonal
Answers
Answer:
hope so it helps you
,......
......
Answer:
Step-by-step explanation:
Given:
A rhombus ABCD
Diagonals AC and BD are perpendicular bisectors
To Prove:
AB² + BC² + CD² + DA² = BD² + AC²
Proof:
In a rhombus we know that the diagonals are perpendicular bisectors.
Hence,
OD = OB = 1/2 BC------(1)
OC = OA = 1/2 AC------(2)
Also in a rhombus, all the 4 sides are equal.
Therefore,
AB = BC = CD = DA------(3)
Now consider ΔODC
By Pythagoras theorem,
DC² = OD² + OC²
Substitute value of OD and OC from equations 1 and 2,
DC² = (1/2 BD)² + (1/2 AC)²
DC² = 1/4 BD² + 1/4 AC²
Multiply the whole equation by 4
4DC² = BD² + AC²
DC² + DC² + DC² + DC² = BD² + AC²
Substitute equation 3 in the above equation,
AB² + BC² + CD² + DA² = BD² + AC²
Hence proved.
\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf A}\put(-0.5,4.2){\bf D}\put(4.2,-0.5){\bf B}\put(4.2,4.2){\bf C}\qbezier(0,0)(0,0)(4,4)\qbezier(4,0)(4,0)(0,4)\put(1.8,1.5){\bf{O}}\end{picture}