03-30-2017, 07:33 PM

(03-30-2017 06:36 PM)Han Wrote: [ -> ](03-30-2017 05:47 PM)Gerson W. Barbosa Wrote: [ -> ]That's what I'd done too, but the solution is very simple, I realize now. No need to solve any linear system.

I suspected there had to be a simple approach. I'll try your hints and also rework my solution to see where I may have made mistakes.

EDIT (after some initial thoughts):

Quote:PS: From E draw a perpendicular line to line AB, which intercepts it at F. Now we have two similar right-triangles: BEF and EFD. Angle DEF = angle EBF = 18 degrees, as we already know. Angle BDE, which you have named 'u' is its complement, 72 degrees. Finally, angle EDC, or your 'x', can be easily determined as 180 - 72 - 48 = 60 degrees.

Are you saying draw EF so that \( \angle BFE \) is 90 degrees? If so, how does it follow that triangle BEF is similar to triangle EFD without also knowing that \( \angle BED \) is 90 degrees (which does not seem obvious to me).

Or are you suggesting we make \( \angle BEF \) equal to 90 degrees? This would make it even less obvious how triangle EFD is even a right triangle.

EDIT 2: Your solution seems to suggest that we only need angles A and C. Am I missing another simple observation ?

I hope the picture is not too much polluted:

Edit: I could have done that ONLY if I were sure BED was a right-triange, which obviously we cannot be sure of.

Back to previous step!