Given the triangle in this diagram, if I draw a line segment (shown in blue) from vertex A to any location on line segment BC, the point of intersection with line segment BC will be at some percentage along the path from vertices B to C.
If I know this percentage along the path from B to C where the blue line intersects line segment BC, how do I calculate the percentage along the path from B to D where the blue line intersects that path? I believe this is a function of three angles: θ, α, and the angle to the blue line?
To be clear: All that is known are the angles θ and α, and the percentage along the path from B to C that the blue line intersects the line segment BC.
Best Answer
Let $E$ and $F$ be the intersection points of the "blue" line with $BC$ and $BD$, respectively.
Assume $\frac{BE}{BC}=p$, $\frac{CD}{CA}=\mu$, $\frac{EF}{EA}=\lambda$.
From $$ \vec{BF}=x\vec{BD} $$ we have: $$ (1-\lambda)\, p\vec{BC}+\lambda\,\vec{BA}=x\left[(1-\mu)\,\vec{BC}+\mu\,\vec{BA}\right]. $$
It follows: $$ \frac{1-\lambda}{1-\mu}p=\frac{\lambda}{\mu} \implies\frac1\lambda=\frac1p\left(\frac1\mu-1\right)+1 $$
so that $x$ in question is: $$ x=\frac\lambda\mu=\frac p{1-\mu(1-p)}.\tag1 $$
It remains only to find $\mu$ from the given angles. Realizing that $\angle CBD=\theta$ one easily finds: $$ \mu=\frac{a\sin\theta}{a\sin\theta+c\sin(B-\theta)}. $$ Substituting $$ B=\frac\pi2-(\alpha-\theta) $$ one finally obtains: $$ \mu=\frac{a\sin\theta}{a\sin\theta+c\cos\alpha},\tag2 $$ which is to substitute into $(1)$.
The conclusion is that $p,\alpha,\theta$ are not sufficient to evaluate $x$. One needs additionally the value of $\frac ca $. It can be easily found if for example the angle $\phi=\angle BAE $ is given.