Question

Prove that diagonals of a rhombus bisect each other?

Answer 1

See the proof below

Explanation:

We prove this with vectors and Chasles' relation

`vec(AC).vec(BD)=(vec(AD)+vec(DC)).(vec(BA)+vec(AD))`

`=vec(AD).vec(BA)+vec(AD)*vec(AD)+vec(DC)*vec(BA)+vec(DC).vec(AD)`

`= AD.BA. cos hat(BAD) +AD^2-AB^2+AD.DC. cos(180-hat(BAD))`

`AD=DC=AB=a`

`cos(180-hat(BAD))=cos180cos hat(BAD)+sin180sin hat(BAD)`

`=-cos hat(BAD)`

Therefore,

`vec(AC).vec(BD)=a^2cos hat(BAD)+a^2-a^2-a^2cos hat(BAD)=0`

So,

As the scalar product of `vec(AC)` and `vec(DB)` is equal to `0`, the sides `AC` and `DB` are orthogonal

`QED`

Answer 2

Proof given below.

Explanation:

Consider the following rhombus `ABCD`, where diagonals `AD` and `BC` intersect at `O`.

In a rhombus all sides are equal and opposite sides are parallel. Further a rhombus is also a parallelgram and hence exhibits properties of a parallelogram and that diagonals of a parallelogram bisect each other.

Hence in `DeltasABO` and `BCO`, we have

`AO=CO` - diagonals of a parallelogram bisect each other

`AB=BC` - sides of a rhombus

and `OB=OB` - common

Hence `DeltaABO-=DeltaBCO`

and `m/_AOB=m/_BOC`

but these two angles are supplementary.

Hence each is a right angle i.e. diagonal of a rhombus are perpendicular to each other.

`Q.E.D.`