I'm attempting to solve this geometry problem:
Image description: Going in a clockwise direction there are upward facing rays $\overrightarrow{BA}$, $\overrightarrow{BE}$, $\overrightarrow{BF}$, $\overrightarrow{BH}$, and $\overrightarrow{BC}$, with ray $\overrightarrow{BA}$ pointed directly to the left and ray $\overrightarrow{BC}$ pointed directly to the right. Angles $\angle ABE$ and $\angle EBF$ are marked congruent. The image states "For 40 and 42. In the figure, ray BA and ray BC are opposite rays. BH bisects angle EBC." The problem is "42. If $m\angle EBC = 3r + 10$ and $m\angle ABE = 2r − 20$, find $m\angle EBF$." It then provides three possible answers: 28, 64, and 32.
My first attempt was:
- $m\angle ABC = m\angle ABE + m\angle EBC = 5r − 10$
- $m\angle ABC = 180^{\circ}$
- $5r − 10 = 180^{\circ}$
- $r = 38$
- $m\angle EBF = m\angle ABE = 2r − 20 = 2\cdot 38 − 20 = 56$
However, 56 was not one of the provided answers. I decided that equating $5r − 10$ and $180^{\circ}$ was invalid, due to not knowing the units of $r$ and also not knowing whether angle values were being measured in radians, degrees, etc. For example, using $\pi$ instead of $180^{\circ}$ gives a completely different answer.
I then determined equations for each possible angle. I'm not sure that ray $\overrightarrow{BH}$ is useful to solving the problem, but I included those angles anyway.
Equations using only $m\angle ABE$:
- $m\angle ABE = 2r − 20$
- $m\angle EBF = m\angle ABE = 2r − 20$
- $m\angle ABF = m\angle ABE + m\angle EBF = 4r − 40$
Equations using only $m\angle EBC$:
- $m\angle EBC = 3r + 10$
- $m\angle EBH = \frac{1}{2}\cdot m\angle EBC = \frac{3}{2}r + 5$
- $m\angle HBC = m\angle EBH = \frac{3}{2}r + 5$
Equations combining $m\angle ABE$ and $m\angle EBC$:
- $m\angle ABC = m\angle ABE + m\angle EBC = 5r − 10$
- $m\angle FBC = m\angle EBC − m\angle ABE = r + 30$
- $m\angle ABH = m\angle ABE + m\angle EBH = \frac{7}{2}r − 15$
- $m\angle FBH = m\angle EBH − m\angle ABE = -\frac{1}{2}r + 25$
In order to find $r$ and then $m\angle EBF$ it seems like I'd first need to determine $m\angle FBH$ or $m\angle FBC$ without using $m\angle EBC = 3r + 10$. However, I'm not seeing a way to do that, so I must be missing something. What is the value of $m\angle EBF$ and how do I determine it?
