justifying reflection across line in beltrami-klein model 
Justify the following construction of the Klein reflection A' of A
  across m.
Let Λ be an end of m and P be the pole of m. Join Λ to A and let this
  line cut y (which is the circle, my note) again at Φ. Join Φ and P and
  let this line cut y at Φ'. Then A' is the intersection of the lines AP
  and ΛΦ'.

[K-6, P.272, Euclidean and Non-Euclidean Geometries - Greenberg, 3rd ed] 
Ok, I'm really stuck at this one but to give you at least an small incentive to help me I've tried;
A' is the harmonic conjugate of A with respect to MP, where P is the pole of line m and M is the point were the line determined by A and P intersects m.
And I know that if m is not a diameter in the Beltrami-Klein model Then the reflection across m is an harmonic homology with center P and with the axis m.
So I need to show that A' is the harmonic conjugate to A with respect to P and M, that is show that the cross ratio (PM, AA') = 1 or equivalently that PA/PA' = MA/MA'.
I tried to use perspectivity with center Λ and came to no conclusion that I found particullarly useful. (wanted to find something equal to (PM, AA') from which => (PM, AA') = 1)
I have also noticed that ΛΦ, ΛΦ' and ΦΦ' are asymptotically parallel and forms a triangle with at least two right angles. 
Anyone who would like to help me out? 
Even the smallest hint would be highly appreciated,
 A: 
Let $\Lambda'$ be the other endpoint of $m$.
The tangents at $\Lambda$ and $\Lambda'$ as well as the line $\Phi\Phi'$ all intersect in $P$.
By Hesse's transfer principle (as explained in Perspectives on Projective Geometry chapter 10.5), this means that $$(\Lambda,\Lambda;\Lambda',\Lambda';\Phi,\Phi')$$ form a quadrilateral set (with respect to the conic, i.e. as seen from any point on the conic).
Excluding exotic degenerate situations, this is equivalent to $$(\Lambda,\Lambda';\Phi,\Phi')$$ being a harmonic quadruple with respect to the circle.
So any point $X$ on the circle will result in a harmonic quadruple of lines $$(X\Lambda,X\Lambda';X\Phi,X\Phi')$$
If you take the limit $X\to\Lambda$, the line $X\Lambda$ will become the tangent $\Lambda P$ and you obtain the harmonic lines $$(\Lambda P,\Lambda\Lambda';\Lambda\Phi,\Lambda\Phi')$$
Now you can choose any points to define these lines, in particular you might as well choose $$(\Lambda P,\Lambda M;\Lambda A,\Lambda A')$$
And since these four points are collinear, the points themselves will be harmonic as well, and you obtain your desired harmonic quadruple $$(P,M;A,A')$$
