Construct a triangle given certain lengths related to a bisector Let $ABC$ be a triangle, and $AD$ the bisector of angle $A$. Write $AB = c$, $AC = b$, $AD = d$, $BD = c'$, $CD = b'$. Using ruler and compass, construct the triangle $ABC$ given the lengths $d$, $b-b'$ and $c - c'$. (That is, we are given segments of lengths $d$, $b-b'$ and $c-c'$ to work with, and the problem is to reconstruct the triangle from these data.)

I already have an analytic solution to the problem that can be translated into a lengthy construction. I am looking for a geometric solution which is as simple as possible.


1) line $AX=x,AY=y,AN=d ,J $ is midpoint of $XY, M $is midpoint of $AX$,
2)circle $J @r=JY,AK \perp AX$, cross circle$J$ at $K$
3)make rectangle$AKQN$, connect $AQ, KL \perp AQ$ , cross $AQ$ at $L$
4) connect $ML,QB$//$ML$ cross $AN$ at $B$
5)circle $B@r=XB$, circle $A@r=AN$, two circle cross at $D$
6)circle $D@r=DN$, cross circle $A@r=AN$ at $P$
7)connect $BD,AP$, cross at $C$
$\triangle ABC$ is the wanted. note $AN> AK$ is necessary condition.
the proof is to OP.
edit: the op comments remind me that there is more simple way to build $AB$

$AG=\dfrac{d^2}{x}+y$
