Prove the diagonals of rhumbus are perpendicular and angles bisector

thank you for your time.

There are many aways to approach this problem, but I tried to do it the long way. I am not quite good at math proofs, which makes me tend to mess the logic up. It would be great if someone verify my proof on this one :)

Let $ABCD$ a rhumbus, then $AB=BC=CD=DA$;

$\triangle ABD$ is isosceles with height $OD$ and base $AC$ (which is one of the diagonals of the rhumbus), therefore $<DAC=<DCA$;

Since $OD$ is the height of $\triangle ADC$, $<DOA = <DOC = 90º$;

$\triangle ABD$ is isosceles with base $BD$ (which one of the diagonal of the rhumbus). Notice that $BD$ intersects $AC$ such that $<AOB$, & $<DOC$, $<COB$ & $<DOA$ are vertically opposite angles, Then $<AOB = <DOC = <DOA=<COB=90º$;

Therefore the diagonals of a rhumbus bisect the angles (because the heights $\in$ the diagonals) and are perpendiculat to the other

HINTS:

Using simple vectors/complex number approach, which is just one step away compared to the Euclidean approach...it calculates easier due to equality all boundary vector lengths.

The rhombus is a particular case of a parallelogram.

$$ e^{i \theta}= \cos {\theta}+ i \sin {\theta} = c+ i\; s = \vec{ BC} = \vec {AD}$$

Perpendicularity/Orthogonality

of diagonals can be verified:

$$ ( ( c+ i s +1) \cdot ( (c +i s -1)= c^2-1+s^2 =0 $$

Arguments of Vectors

$$ \arg( 2 + e^{i \theta}) = \theta$$

$$ \arg( 1 + e^{i \theta} )= \theta + \theta =2 \theta $$

( External angle sum equals sum of two internal angles)

The resultant vector is hence a bisector also.