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.


I assume that $\angle BAC$ is also given. Let $AX = x = b – b’$ and $AY = y = c – c’$. By angle bisector theorem, x : y = b’ : c’.
(1). Construct the circle $\alpha$ (centered at A, radius = d).
(2) Locate $B_1$ (exterior to ⊿ABC) on $\alpha$ such that $\angle B_1AB = \angle BAD$. $C_1$ is similarly constructed.
(3) B and C will be somewhere on the perpendicular bisectors of $B_1D$ and $C_1D$ respectively.
(4) B and C will be somewhere on the perpendicular bisectors of $B_1Y$ and $C_1X$ respectively.
(3) and (4) determine length and location of the line BC.
