Given $F_1$, $F_2$ are foci of hyperbola $h$, $A$ is a point on the second branch of $h$, $AB$ is a bisector of $F_1AF_2$, $\alpha = \angle F_1AB=\angle BAF_2$, $\beta=\angle ABF_2$, proof that: $$ \frac{\cos\alpha}{\cos\beta} = \epsilon, $$ where $\epsilon$ is eccentricity.
I am interested in a simple geometric proof.
Below is the physics/algebraic proof I came up with. It can be made rigorous if needed.
- Take a neighbour point $A'\in h$, $AA' = dx \ll AF_{1,2}$, draw perpendiculars $A'H_1$, $A'H_2$ to $AF_1$ and $AF_2$
- Since $\angle AF_1B = \beta-\alpha$ and $\angle AF_2B = \pi-\beta-\alpha$, , we conclude
$$A'H_1 = AF_1 (d\beta-d\alpha), \qquad A'H_2 = AF_2(d\beta+d\alpha)$$
From $A'H_1=A'H_2 = dx \sin\alpha$, we find $$ \frac{d\alpha}{d\beta} = \frac{AF_1-AF_2}{AF_1+AF_2} $$
From sine law in $ABF_2$ and bisector property ($F_1A:F_2A = F_1B:F_2B$): $$ \frac{\sin\alpha}{\sin\beta} = \frac{BF_2}{AF_2} = \frac{F_1F_2}{AF_1+AF_2} $$
From (3) and (4): $$ \frac{d\cos\alpha}{d\cos\beta} = \frac{\sin\alpha\ d\alpha}{\sin\beta\ d\beta} = \frac{AF_1-AF_2}{F_1F_2} = \frac{2a}{2c}=\epsilon $$
Thus, $\cos \alpha = \epsilon\cos\beta + C$, and when $A\to$ point on $F_1F_2$, $\cos\alpha\to 0$ and $\cos\beta\to 0$, so $C=0$
Apply the cosine rule to triangles $AF_1B$ and $AF_2B$, to find $BF_1^2$ and $BF_2^2$. Subtracting the results we get: $$ BF_1^2-BF_2^2=AF_1^2-AF_2^2-2AB(AF_1-AF_2)\cos\alpha. $$
Apply then the cosine rule to the same triangles, to find $AF_1^2$ and $AF_2^2$. Subtracting the results we get: $$ AF_1^2-AF_2^2=BF_1^2-BF_2^2+2AB(BF_1+BF_2)\cos\beta. $$ Comparison of these equalities yields $$ 2AB(BF_1+BF_2)\cos\beta=2AB(AF_1-AF_2)\cos\alpha, $$ that is: $$ {\cos\alpha\over\cos\beta}={BF_1+BF_2\over AF_1-AF_2}=\epsilon. $$