Let ABCD be a rectangle with DA=12.
ABJ is an equilateral triangle with J inside ABCD.
CHI and EGF are congruent equilateral triangles with E and F on DA, H on CB, J on FG, and E, I, G, H collinear.
(In reply to My attempt
I just posted my solution.
Yours has some errors:
J is not necessarily on line as EH. (But it can be - you'd need to show this. Then I and J coincide and the solution is simple.)
JB is not equal to EF. Its the shorter side of BJH that's equal. (The solution becomes simple.)
Posted by Jer
on 2020-03-14 17:39:47