BAO : Relation between redshift, Hubble constant and radial

From this link https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations#Measured_observables_of_dark_energy , I can't get this relation :

$$c\Delta z = H(z)\Delta \chi\quad\quad(1)$$

with $$z$$ redshift, $$H(z)$$ Hubble constant at redshift = $$z$$ and $$\chi$$ radial coordinates.

One starts from angle of object $$\Delta \theta$$ which is equal to the ration :

$$\Delta\theta = \dfrac{\Delta\chi}{\text{d}_{a}(z)}\quad\quad(2)$$

with $$\text{d}_{a}(z)$$ the angular diameter distance at redshift=$$z$$.

It is indicated also on this page the relation for angular diameter distance $$\text{d}_a(z)$$:

$$\text{d}_a(z)\propto \int_{0}^{z}\dfrac{\text{d}z'}{H(z')}\quad\quad(3)$$

Actually, I know that $$\text{d}_a(z)$$ is expressed as a function of cosmological horizon $$\text{d}_{h}(z)$$ and redshift $$z$$ like this :

$$\text{d}_{a}=\dfrac{\text{d}_{h}(z)}{1+z}\quad\quad(4)$$

with $$\text{d}_{h}(z)=c\int_{0}^{z}\dfrac{\text{d}z}{H(z)}\quad\quad(5)$$

So from $$eq(5)$$, what I can only write is (by considering a little $$\Delta$$ and a curvature parameter $$\Omega_{k}=0$$) :

$$c\Delta z=\text{d}_{h}(z)H(z)\quad\quad(6)$$

Now, taking the expression of $$\text{d}_{h}(z)$$ into $$eq(6)$$ :

$$c\Delta z=\text{d}_{a}(z)(1+z)H(z)\quad\quad(7)$$

Then :

$$c\Delta z=\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)\quad\quad(8)$$

As you can see in $$eq(8)$$, this is not the same form as in $$eq(1)$$.

How can I make disappear the factor $$(1+z) /\Delta\theta$$ in order to have simply for the right member : $$H(z)\Delta \chi$$ instead of $$\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)$$ into $$eq(8)$$ ?

$$\Delta\chi$$ represents for me the variation $$\Delta$$ of radial coordinate, doesn't it ?

CORRECTION 1 :

As suggested by Michelle Grosso, the eq(3) is not the definition of for angular diameter distance $$\text{d}_{a}(z)$$ but the definition of radial coordinate :

$$\chi = \int_{0}^{z}\dfrac{c\text{d}t}{R(t)}$$

Taking $$\text{d}t=\dfrac{\text{d}t}{\text{d}z} dz$$ with $$1+z=\dfrac{R_{0}}{R(t)}$$, so we get :

$$\dfrac{\text{d}z}{\text{d}t} = -\dfrac{R_{0}\,\dot{R(t)}}{R(t)^2} = -H(z)(1+z)$$

$$\Longrightarrow\quad \dfrac{\text{d}t}{\text{d}z} = -\dfrac{1}{(H(z)(1+z)}$$

$$\Longrightarrow\quad \chi = \int_{0}^{z}\dfrac{c\text{d}z}{(1+z)\,R(t)\,H(z)}$$

$$\chi = \int_{0}^{z}\dfrac{c\,\text{d}z}{R_{0}\,H(z)}$$

we have finally :

$$c\Delta z = H(z)\,R_{0}\,\Delta\chi\quad\quad(1)$$

Is is a better calculus ?

The angular diameter distance is defined as $$d_A = \Delta S / \Delta \theta$$ where $$\Delta S$$ is the proper transverse size of an object at redshift $$z$$ and $$\Delta \theta$$ is its observed angular diameter.

Eq(2) is not correct as you posted as proper transverse size $$\Delta \chi$$, which is instead the variation of the radial coordinate.

Eq(1) can be worked out via the RW (Robertson-Walker) metric written as
$$ds^2 = -dt^2 +a^2(t) R_0^2 [d\chi^2 + S_k^2(\chi) d\Omega^2]$$
where:
$$c = G = 1$$ natural units
$$a(t)$$ scale factor (dimensionless)
$$R_0$$ radius of the universe as today $$(t = t_0)$$
$$\chi$$ radial coordinate
$$S_k(\chi) =$$
$$sin(\chi), k = 1$$ positive curvature (closed universe)
$$\chi, k = 0$$ no curvature (flat universe)
$$sinh(\chi), k = -1$$ negative curvature (open universe)
$$d\Omega^2 = d\theta^2 + sin^2\theta d\phi^2$$ metric on the two-sphere
Today $$a(t) = a(t_0) = a_0 = 1$$

On a null geodesic (photon), chosen radial for convenience we have
$$0 = ds^2 = -dt^2 + a^2 R_0^2 d\chi^2$$
$$d\chi = R_0^{-1} \frac{dt}{a} = R_0^{-1} \frac{da}{a^2 H(a)}$$
where:
$$H = \dot a / a = (da/dt) / a$$ Hubble parameter

Converting the scale factor to redshift via $$a = 1 / (1 + z)$$ we have
$$d\chi = R_0^{-1} \frac{dz}{H(z)}$$
This is your eq(1). Just note that I used a dimensionless radial coordinate $$\chi$$.

Note: Eq(3) is not the angular diameter distance. It is the radial coordinate distance.