Bondi's k factor Consider two observers, travelling away from each other, after meeting (at which time they sync their clocks). $O$ sends a photon towards $O'$ at times $t$, received by $O'$ at time $t'=kt$, where we have defined Bondi's k factor.
In the following reasoning I know I have made a mistake: So by Milne's radar definition of simultaneity, if the photon is instantly sent back from $O'$ and received by $O$ at $t_2$, then:
$$t'=\frac12(t+t_2)=kt\Rightarrow t_2=2t'-t=(2-\frac{1}{k})t'$$
But $t_2$ is meant to equal $kt'$
Could someone explain to me where I have I gone wrong here?
EDIT: Basically I am trying to show that $t_2=kt'$ working from only the two assumptions that:
(i) they both observe the speed of light as c
(ii) Only relative motion is observable

 A: Your formula
$$ t' = \frac{t+t_2}{2} $$
implicitly uses the time measured by the observer with the vertical world line – which is the same thing as the vertical $y$-coordinate. However, the tilted/moving observer measures a shorter time $t'$ by the time dilation factor $\sqrt{1-v^2/c^2}$.
I know that my method of calculation isn't following the pedagogical goal of Bondi to convert everything to the Bondi's $k$-factor, the Doppler factor
$$ k =  \sqrt{\frac{1+v/c}{1-v/c} }$$
but I haven't learned relativity in Bondi's way and the main goal is to fix the discrepancy which should be allowed to be done in any way. So let me say that the time dilation occurs by the factor $\sqrt{1-v^2/c^2}\lt 1$. So let's revert the relationships. For the $k$ defined above, we have
$$ \frac vc = \frac{k^2-1}{k^2+1},\quad \sqrt{1-\frac{v^2}{c^2}} = \frac{2k}{k^2+1} $$
This is the time dilation factor you have omitted so the right equation replacing 
$$ t' = \frac{t+t_2}{2} $$
is 
$$ t' = \frac{k}{k^2+1} (t+t_2) $$
This is right. Substitute $t_2=kt'$ and you get
$$ t' = \frac{kt}{k^2+1} + \frac{k^2}{k^2+1}t' $$
i.e., after multiplication by $k^2+1$,
$$ t' (k^2+1-k^2) = kt,\quad  t'=kt$$
which you wanted to get.
Inverted calculation
For Freeman, let me also revert the logic and order of the calculation. By the definition of Bondi's $k$, we may assume $t'=kt$. We want to calculate the time dilation factor $1/\gamma \lt 1$ – I want to avoid new symbols. The time $t'=kt$ may also be expressed as
$$ t' = \frac{1}{\gamma} \frac{t+t_2}{2} $$
i.e.
$$ kt = \frac{1}{\gamma}\frac{t+k^2 t}{2} $$
where I used $t_2=kt'=k^2 t$. From the last equation, it is easy to calculate
$$\frac{1}{\gamma} = \frac{2k}{k^2+1},$$
just like expected.  
A: 
Is it due to this similar triangle relation, derived from the fact light rays will always be seen at the same speed, so we draw them at $\pi/4$
(larger link http://i.stack.imgur.com/ZaVOd.jpg)
