where I hope to answer the posed problem in a coherent and clear way. So let’s begin.
Breaking the Rules
As I discussed in my earlier post the first solution I was able to get was using Trigonometry, specifically the Law of Sines. I made many attempts at this before I found a solution, and it was my second time threw solving this problem that I developed a cleaner solution. I will walk through this process through a series of pictures shown below:
After careful observation (found from many attempts at this problem) the ratio to consider is found from triangle ADE using the Law of Sines.
Forming the ratios as seen in the above drawing using the two triangles as shown yields the desired ratio which is below.
Setting the two ratios equal will give the resulting equation we need to determine the roots, which gives the solution to then unknown quantity. I was very proud of this solution as it is much cleaner than the two pages of dense writing and equations that I had produced on my first attempt. Now I will write the solution using only elementary geometry.
Following the Rules
To begin, note that the base angles of the triangle ABC are both the sum of two angles whose value is 80 degrees, so triangle ABC is an isosceles triangle by the converse of the Base Angle Theorem. My next step is to construct a line parallel to the line segment AB going through the point D, I labeled the intersection the parallel line through D and the side CB as the point F.
The construction of the parallel line allows us to write that the triangle ABC is similar to triangle DFC by the AA Theorem. We can also determine a variety of angles using properties of parallel lines, like corresponding angles are congruent and alternate interior angles are congruent, which will play an important role in our next step showing the triangle DBF is congruent to triangle FAD. The triangle FAD is built by connecting the points F and A with a line.
To show that the two triangles are congruent, we know that angle ADF and angle DFB both measure 100 degrees, so they are congruent. Likewise, by the reflexive property of congruency we know that the segment DF is congruent to FD. Finally, we know that both angles BDF and AFD measure 60 degrees. This allows us to say that triangle ADF is congruent to triangle BDF by angle-side-angle (ASA) postulate.
In proving the two triangles are congruent, we find that there are also two equilateral triangles, triangle DFG and triangle AGB, which gives multiple congruencies between the corresponding legs. At this point, we see if we can show that the segment DF is congruent to the segment EF, then the triangle DFE is an isosceles triangle and we would be able to recover the solution to the unknown quantity. We already know that the segment DF is congruent to both segments DG and FG respectively since triangle DFG is an equilateral triangle since it is an equiangular triangle.
Next we note that the triangle CAF is an isosceles triangle with congruent base angles of 20 degrees. So we know then that the segment AF is congruent to the segment CF. We would like to find a relationship using the congruency established with the isosceles triangle CAF. As it stands though, I could not find a relationship that would establish a means to an ends.
A helpful hint at this point is to construct an angle bisector through angle ACB which connects to the point G. The resulting triangles formed yields the desired connection we were looking for. Specifically, we see that we want to show that the triangle CAE is congruent to the triangle CAG.
To show the two triangles are congruent, we note that angle ACG is congruent to angle CAE since both measure 10 degrees. In addition, notice that AC is congruent to CA by the reflexive property of congruency. Finally, we have angle CAG is congruent to ACE since both share the measure of 20 degrees. So, by ASA we have the two triangles ACG and ACE are congruent.
Establishing this congruency is very important to our task of relating the length of DF and EF. We can now write the segment AG is congruent to the segment CE, but we also know that the segment AF is congruent to the segment CF, so if CE is congruent to AG, then by congruency of lengths we know that GF is congruent to EF.
Since GF is is congruent to EF, then we know that DF is congruent to GF which is congruent to DG which is congruent to EF. This allows us to find the value of x using the fact that DF is congruent to EF, we have an isosceles triangle. The vertex angle of the triangle is 80 degrees and the two congruent base angles is the sum of the unknown quantity and 30 degrees.
Setting up the equation, we obtain our desired result as seen in the figure, verifying our result from the trigonometric approach above.
I need to give credit where credit is due, so the solution to this problem using geometry is my own in the sense of that just described, but I relied heavily on the following website, while initially wrapping my brain around the details. The sheets that I printed out and worked on were also from this website, which is where I got two of the three problems in this section. I will be posting those problems and their solutions in a similar way, provided folks like the format.