"World's Hardest Easy Geometry Problem"

Just make use of elementary geometry you say? As you wish...

All the angles in a triangle add up to 180º, so I can fill in some blanks on the picture.

$m\angle ACB = 20º$

$m\angle ADB = 40º$

$m\angle AEB = 30º$

There's more to find, but it leads to a dead end...

So I just need more lines on this picture (crazy, right?). As of this point in geometry, I would know about parallel lines and transversals. So to start, I need to draw the line parallel to $\overline{AB}$ through $D$ , intersecting $\overline{BC}$ at $F$ (so $\overline{AB} \parallel \overline{DF}$ ). Then I can draw $\overline{AF}$ , intersecting $\overline{BD}$ at $G$ . Finally, I can draw $\overline{CG}$ .

Now, there are quite a few congruent triangles on this picture, but none that can be proven with the information I have thus far. But I can fill in more blanks.

Corresponding Angles are Congruent, so:

$m\angle CAB = m\angle CDF = 80º\ \color{#3D99F6}{\Rightarrow} m\angle BDF = 60º$

$m\angle CBA = m\angle CFD = 80º$

There are now several isosceles triangles on this picture, two of which are $\triangle ABC$ and $\triangle DFC$ . So,

$\overline{DC}\cong\overline{FC}$

$\overline{AC}\cong\overline{BC}$

Subtract these two: $AC - DC = BC - FC \color{#3D99F6}{\Rightarrow} \overline{AD}\cong\overline{BF}$

I now have everything necessary to say that by the SAS postulate, $\triangle ABD \cong \triangle BAF$ . Thus,

$m\angle DBA = m\angle FAB = 60º\ \color{#3D99F6}{\Rightarrow} m\angle FAE = 10º$

$m\angle ADB = m\angle BFA = 40º\ \color{#3D99F6}{\Rightarrow} m\angle AFD = 60º$

That is the second angle of $\triangle DFG$ , thus $m\angle DGF = 60º$ , which makes $\triangle DFG$ equiangular, and thus equilateral. Thus,

$\overline{DF}\cong\overline{DG}\cong\overline{GF}$

I now have everything necessary to say that by the SSS postualte, $\triangle CGD \cong \triangle CGF$ . Thus,

$m\angle ACG = m\angle BCG = 10º$

And now a tricky one to see, I now have everything necessary to say that by the ASA postulate, $\triangle ACG \cong \triangle CAE$ . Thus,

$\overline{AG}\cong\overline{EC}$

$\triangle AFC$ is isosceles, so $\overline{AF}\cong\overline{FC}$

Subtract those two: $AF - AG = FC - EC \color{#3D99F6}{\Rightarrow} \overline{GF}\cong\overline{EF} \color{#3D99F6}{\Rightarrow} \overline{DF}\cong\overline{EF}$

That makes $\triangle DFE$ isosceles. Therefore,

$2(x+30)º + 80º = 180º$

$\color{#3D99F6}{\Rightarrow} \color{#D61F06}{x = 20º}\space\space\space\Box$

Use the law of cosines to solve for cos(x) in terms of the base and sides of the large isosceles triangle. Two equations in two unknowns

My triangle is this one, I know the "correct" answer is 20, but this is what I did and my answer is x = 50°. Please let me know if you find a mistake so I can correct it, thank you

https://www.facebook.com/photo.php?fbid=994470413942408&set=a.610002989055821.1073741834.100001383774807&type=3&theater

A similar difficult problem will be with the same drawing (not to scale) but with different angles: the pair 70 - 10 becomes 60 - 20 and the pair 60 - 20 becomes 50 - 30

Draw AB।।DG and DG।।EF. So, <FEG=100. (as, <CEF=80). DE divide <FEG equally. So,<FED=<DEG=50. <DEG=x+30=50. So, x=50-30=20.