First, we need some auxiliary lines and points. Draw a line through D parallel to AB, and let F be the intersection with BC. Now draw line AF. (It's just a mirror image of BD.) Let G be the intersection between AF and BD. Finally, draw line CG. It's easy to see that the line CG bisects the angle ACB, so ACG and GCB are both 10 degrees.

It's helpful to keep the endgame in mind while working toward it. Just from the fact that the angles of a triangle add up to 180 degrees, We know that angle AEB is 30 degrees. If we knew angle DEF, we could easily compute x: DEF - AEB. So, how to find angle DEF? First we need to establish three more-or-less unrelated equations between segment lengths:

- DF = FG. Angles ABG and BAG are both 60 degrees, so ABG is an equilateral triangle. Since angle DGF = AGB, angle DGF is also 60 degrees. By symmetry, angles FDG and DFG are equal, so they must be 60 degrees, too, and DFG must be equilateral. So the sides are equal.
- CF = AF. Since angles FCA and FAC are both 20 degrees, ACF is isosceles, which means that CF = AF.
- CE = AG. This is the trickiest one. It hinges on the fact that triangles ACG and ACE have equal sides. That's a little surprising, but true: both have angles 10, 20, and 150, and they have a common side AC. So they're congruent and have a common side, so they're equal. That means in particular that CE = AG.

- CF - CE = AF - AG

- EF = FG

## 5 comments:

O MY GOD! Thanks sooo much! I was at my wits end trying to solve this!

Now I can get back to work...

is it bad that I'm a second year maths student at uni and this problem, despite the proof, is still confusing me :S

Couldn't get your tricky statement...

not a good proof. easier one is at

http://blogs.oracle.com/simford/entry/solved_the_world_s_hardest

solved it in 3mins just after listening (i'm 14)

Post a Comment