Question 474056
{{{drawing(300,300,0,10,0,10,


triangle(1,1,9,1,4,8),
locate(1,1,B),
locate(9,1,C),
locate(4,8,A),
line(4,8,4.8,1),
locate(4.8,1,D),
locate(6.5,4.5,E),
locate(3.5,7,alpha),
locate(4.5,7,alpha),
line(4.8,1,6.5,4.5),
locate(4.8,2.3,alpha)

)
}}}


First, we note that triangle ADE is isosceles, so angle ADE equals alpha. Now we can say that lines DE and BA are parallel, so triangles CDE and CBA are similar (by an AAA argument).


Because they're similar, we can establish


*[tex \LARGE \frac{CD}{CB} = \frac{CE}{CA}]


Rewrite CD as CB - BD and CE as CA - AE.


*[tex \LARGE \frac{CB - BD}{CB} = \frac{CA - AE}{CA}]


*[tex \LARGE 1 - \frac{BD}{CB} = 1 - \frac{AE}{CA} \Rightarrow \frac{BD}{BC} = \frac{AE}{AC}], as desired. Here I switched the order of the points (e.g. BC to CB) because each segment refers to the magnitude of the line segment. There are some theorems where line BC might refer to a vector quantity, in which BC = -CB, and switching the order of the points would change everything.