I'm working on a problem in What is Mathematics? by Richard Courant and Herbert Robbins. The problem statement is : "If in a set of four harmonic lines a,b,c,d, the ray a bisects the angle between c and d, then b is perpendicular to a."
Based on the problem description I begin with this image: 
I've taken two approaches so far.
Approach 1:
Because the set of lines is harmonic the cross ratio of (a,b,c,d) = -1. Next, my book states $(1,2,3,4)= \frac{\frac{\sin{1,3}}{\sin{2,3}}}{\frac{\sin{1,4}}{\sin{2,4}}}$.
Using (a,b,c,d) in place of (1,2,3,4) and letting the bisected angle $\angle cpd = x$ and $\angle dpd = y$ I get:
$$-1 = \frac{\frac{\sin{\frac{-x}{2}}}{\sin{-x-y}}}{\frac{\sin{\frac{x}{2}}}{\sin{-y}}}$$
Which I show reduces to $\sin(x+y) = \sin(y)$. But that is clearly the wrong answer.
Approach 2: Because the set of lines is harmonic the cross ratio of (a,b,c,d) = -1. However because the image shows them in the order (c,a,d,b) we need to use $1-\lambda = 1-(-1) = 2$.
Using (c,a,d,b) in place of (1,2,3,4) and letting the bisected angle $\angle cpd = x$ and $\angle dpd = y$ I get :
$$-1 = \frac{\frac{\sin{x}}{\sin{\frac{x}{2}}}}{\frac{\sin{x+y}}{\sin{\frac{x}{2}+y}}}$$
But this also give me that x = y = 0.
Questions
- Based on my errors I think I'm setting up the initial equations wrong. Specifically, I don't think I should have $\sin(x+y)$ but if I don't use $\sin(x+y)$ what should my cross ratio formula be?
- If I am supposed to have $\sin(x+y)$ where does my process go wrong? I believe I'm supposed to end up with $\sin(x) = \sin(y)$ not $\sin(x+y) = \sin(y)$.
Draw any line parallel to $b$ and let the intersection of this line with $a$, $b$, $c$ and $d$ be $A$, $B_{\infty}$, $C$, $D$, respectively. Then $(CDAB_{\infty})$ is harmonic, so $A$ is the midpoint of $CD$. If $O$ is the intersection of your lines, then the triangle $OCD$ is isosceles, since $OA$ is angle bisector and median as well. So $OA$ is perpendicular to $CD$, which is parallel to $b$.