Let's say I have a triangle ABC, with side lengths abc. I need to draw a line from the angle connecting the base (c) and hypotenuse (b). I don't know the real angle, but I know it's sin-1. I need to draw a line from that angle, to the opposite side (BC), but the angle at which the line is drawn must be half the angle mentioned, therefore 0.5(sin^-1 (a/c) ). Is there an expression that tells me where to put the endpoint of this line (on BC)?
[[BTW: Please give answers in terms of or using ABC as vertices, and abc as side lengths of the two remaining vertices combined. aka: a is the side length of BC. b is the side lenght of AC. c is the side length of AB. Case matters!]]
One more question (might be repetitive): I need the length of the line AD (or A>Endpoint). How do I get this length AD?
By the Angle Bisector theorem, we have the relationship $\dfrac{BD}{CD} = \dfrac{AB}{AC} = \dfrac{c}{b}$, where $D$ is the point on $BC$ such that $\overrightarrow{AD}$ bisects $\angle BAC$.
Add $1$ to both sides to get $\dfrac{BD}{CD}+1 = \dfrac{c}{b}+1$, i.e. $\dfrac{BD+CD}{CD} = \dfrac{b+c}{b}$.
Since $BD+CD = BC = a$, we get $\dfrac{a}{CD} = \dfrac{b+c}{b}$. Solve for $CD$ to get $CD = \dfrac{ab}{b+c}$.
This is enough information to tell you exactly where to place $D$ on $BC$.
You can plug this back into $BD+CD = a$ to get $BD = BC-CD = a-\dfrac{ab}{b+c} = \dfrac{ac}{b+c}$.
To get $AD$, apply Stewart's Theorem as done here to get $AD = \sqrt{bc\left(1-\dfrac{a^2}{(b+c)^2}\right)}$
Note: The above formulas hold even if $\angle ACB \neq 90^{\circ}$.