Prove that the diagonals of a rhombus are at right angles when O is origin c ( α ,β) B ( α + h ,β) A (h, 0)?

Question

1385 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-12T13:20:27

Please see the proof below

Let's work with vectors

$vec(OA)=((h),(0))$

$vec(CB)=((h),(0))$

$vec(OC)=((alpha),(beta))$

$vec(AB)=((alpha),(beta))$

As the figure is a rhombus

$||vec(OA)||=||vec(CB)||=||vec(OC)||=||vec(AB)||$

$h^2=alpha^2+beta^2$

The dot product of the diagonals is

$vec(OB). vec(AC)=((alpha+h),(beta)).((alpha-h),(beta))$

$=(alpha+h)(alpha-h)+beta^2$

$=alpha^2-h^2+beta^2$

$=0$

Therefore,

The vectors $vec(OB)$ and $vec(AC)$ are orthogonal, that is
the diagonals are perpendicular to one another.