For the obtuse ∠ABC, right ∠ADE, and FG where F is on AB and G is on BC what is the length of DE, where E is the bisector of FG.
The title technically explains it, but I've drawn a diagram to make it easier to understand.
Diagram of the problem (sorry for the poor quality).
I need two pieces of information from the diagram (I'm using this for a coding project, but that is only relevant because it explains why I don't use any numbers after this. All the values are dynamic). I need to find the measure of $\overline{DE}$ and the measure of ∠EFD (and I know I can find one once I have the other). Here's the information I have to start:
- The slopes of $\overline{AB}$ and $\overline{BC}$ in $\frac{rise}{run}$
- The measure of ∠ABC is > 90
- ∠ADE is a right angle
- The length of $\overline{FG}$
- E bisects $\overline{FG}$
Using math, I know how to get:
- The measure of ∠ABC
- The measure of $\overline{FE}$ and $\overline{EG}$
This is where I get stuck. I have a bunch of information, but I can't figure out how to put it together. Thanks in advance for the help!
Note: Since I posted this question, I noticed several things that cause it to not be a full solution to the problem I need it for. Thus, I have posted an updated version. If you want to help me, here's the link: https://stackoverflow.com/questions/76498751/finding-footing-for-proceedural-animation
It is necessary to consider two slightly different cases here. What you see below is the solution for the case depicted in $\mathrm{Fig.\space 1}$, in which the point $D$ lies between the two points $F$ and $B$. We leave OP to work out a solution for the case shown in $\mathrm{Fig.\space 2}$, in which the point $B$ lies between the two points $F$ and $D$.
Please note that all known lengths and angles are shown in $\bf{\color{red}{red}}$. We have extended the segment $AB$ and dropped a perpendicular to from $G$ to meet it at $H$. Furthermore, for brevity, we let $\angle GBH = \theta$ and $\angle GFB=\beta$. We start by applying the midpoint theorem to the right-angled triangle $FHG$ to obtain, $$DE=\dfrac{GH}{2}=\dfrac{BG\sin\left(\theta\right)}{2} \qquad\text{and}\tag{1}$$ $$FD=\dfrac{FH}{2}=DB+BG\cos\left(\theta\right).\quad \tag{2}$$
When we apply the low of cosiness to the obtuse triangle $FBG$, we have, $$FG^2=FB^2+BG^2+2FB\times BG\cos\left(\theta\right). \tag{3}$$
Now, using (2), we can express $FB$ as shown below. $$FB=FD+DB=2DB+ BG\cos\left(\theta\right) \tag{4}$$
After replacing $FB$ in (3) using (4), we simplify (3) to obtain the following quadratic equation in $BG$. $$\left(1+3\cos^2\left(\theta\right) \right)BG^2+\left(8DB\cos\left(\theta\right)\right)BG + 4DB^2-FG^2=0 \tag{5}$$
This equation has two real roots, but only one of them is positive, i.e., $$BG=\dfrac{\sqrt{\left(1+3\cos^2\left(\theta\right) \right)FG^2-4DB\sin^2\left(\theta\right)}-4DB\cos\left(\theta\right) }{1+3\cos^2\left(\theta\right) }. \tag{6}$$
We can use (1) and (6) to obtain the sought expression for the determination of length of $DE$. $$DE=\dfrac{\sqrt{\left(1+3\cos^2\left(\theta\right) \right)FG^2-4DB\sin^2\left(\theta\right)}-4DB\cos\left(\theta\right) }{2\left(1+3\cos^2\left(\theta\right)\right)} \sin\left(\theta\right) $$
Finally, by considering the right angle triangle $FDE$, we can derive a formula to find the value of $\beta$. $$\beta {\small{=\dfrac{DE}{FE}=\dfrac{2DE}{FG}=}} \sin^{-1} \left(=\dfrac{\sqrt{\left(1+3\cos^2\left(\theta\right) \right)FG^2-4DB\sin^2\left(\theta\right)}-4DB\cos\left(\theta\right) }{\left(1+3\cos^2\left(\theta\right)\right)FG }\sin\left(\theta\right)\right)$$