Begin with unit equilateral triangle ABC.
Extend side AC one unit to D so DC=1.
Construct line AE such that
(1)E is on line DB with B between D and E,
(2)line AE intersects CB at F with B between C and F,
(3)and EF=1.
Find the length AE.
Comment on the title.
since ABC is equilateral that means angle ACB=60 and thus BCD=120.
Since BC=CD=1 we have that angle CBD = angle CDB = 30
Since E,B, and D are colinear then we have angle EBA=90
from that since F,B, and C are colinear then we have angle FBE=30
let x be the measure of the angle EAB then we have side EB = tan(x)
angle AEB=90x
angle FEB=90+x
angle EFB=60x
using the law of sines in triangle EFB using angles EFB and FBE we have
sin(30)=sin(60x)/tan(x)
sin(30)*tan(x)=sin(60x)
(1/2)*tan(x)=sin(60x)
tan(x)=2sin(60x)
now I don't see any way of solving this other than numerically for which I get that x is approximately 0.653928
which then gives that AE=1/cos(x)=1.25992
Not sure how this relates to the title which gives me the feeling that there is a shortcut relating to the title.

Posted by Daniel
on 20170604 14:27:49 