Get ∠EBF=∠ABC, that is, EB=EF. Let caf = b, then BAF = a-b, Abe = BEF-BAF = a-(a-b) = b.
From the area of △ABD: △ACD =BD:DC=k, we can know from the triangle area formula.
(ab * ad * sin(a-b)/2): (AC * ad * sinb/2) = k, so sin(a-b):sinb=k, and then expand sin (a-b).
Sinacosb-sinbcosa=ksinb, so we know that tanb=sina/(k+cosa).
Consider that in △ACF, ∠AFC=∠ABC=(π-a)/2, so ∠ACF=(π+a-2b)/2.
Let ∠CEF=x, then ∠FCE=(π+a-2x)/2, which can be obtained by sine theorem.
CF:EF = sinx:sin((π+a-2x)/2)= sinx:cos(x-a/2).........①
CF:AF = sinb:sin((π+a-2b)/2)= sinb:cos(b-a/2).........②
And Fe: ea = be: ea = sin ∠ BAF: sin ∠ Abe = sin (a-b): sinb = k, so FE:AF=k:(k+ 1). Combining ① and ②, it can be concluded that.
Ksin x: cos (x-a/2) = (k+1) sinb: cos (b-a/2), that is, ksinxcos (b-a/2) = (k+1) sinbcos (x-a/2),
Expand both sides of cos(x-a/2) and cos(b-a/2) and divide them by cosbcosx, and you can get it.
KTANX (COS (a/2)+TANBSIN (a/2)) = (k+1) TANB (COS (a/2)+TANXSIN (a/2)), from which we can find out.
tanx =[(k+ 1)tanb cos(a/2)]/[kcos(a/2)-tanb sin(a/2)]
Then bring in tanb=sina/(k+cosa) and cut off cos(a/2)/(k+cosa) at the same time, and you can get it.
tanx =(k+ 1)Sina/(k(k+cosa)-2 sin？ (a/2)), and reuse 2sin? (a/2)= 1-cosa, just bring it in.
tanx=(k+ 1)sina/(k？ - 1+(k+ 1)cosa)= Sina/(k- 1+cosa)