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.
 A: 
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 $AB$

$AG=\dfrac{d^2}{x}+y$
A: 
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.
A: a more nice method to rebuild the triangle:
 
$AD=d,AK=c-c',AJ=b-b', F $ is mid point of $AD$
1) circle $F@r=AF$,circle $A@r=AK$,circle $A@r=AJ$, three circles cross at $K_1,J_1$
2)$PQ \perp AD, AK_1$ cross $PQ$ at $P$,cross circle $AJ$ at $J_2, AJ_1$ cross $PQ$  at $Q$, cross circle $AK$ at $K_2$ 
3) take midpoint $M$ of $QK_2$, midpoint $N$ of $PJ_2$, connect $MN$ ,cross $AD$ at $I$
4)circle $I@r=ID$, cross circle $AK$ at $K_3$,cross circle$AJ$ at $J_3$
5)circle $A@r=AM,A@r=AN$, connect $AK_3$ cross circle $AM$ at $B$, connect $AJ_3$ , cross circle $AN$ at $C$.
$\triangle ABC$ (or $\triangle AB_1C_1$) is the one wanted.
A: Assume $b > c$ (hence also $b' > c'$ and $b - b' > c - c'$, since $b'/b = c'/c$).
Let $I$ be the incenter of $ABC$, and $J$, $K$, $L$, its projections onto $BC$, $AC$ and $AB$ respectively. Write $e = AK = AL$, $f = DJ$, $d_1 = AI$, $d_2 = ID$. 
Then I claim (proofs given below) that:


*

*$e = \frac{1}{2}[(b - b') + (c - c')]$,

*$f = \frac{1}{2}[(b - b') - (c - c')]$,

*$d_1 + d_2 = d$,

*$d_1^2 - d_2^2 = e^2 - f^2 = (b - b')(c - c')$.


Construction. We first construct the length $e$ in the obvious way, and then $f = e - (c - c')$. Now we construct a right triangle with hypotenuse $e$ and one leg $f$, hence its other leg is $g = \sqrt{e^2 - f^2} = \sqrt{d_1^2 - d_2^2}$.
To construct the length $d_1$, we draw a right triangle with legs $d$ and $g$. The intersection of the bisector of the hypotenuse and the leg of length $d$ divides that leg into segments of length $d_1$ and $d_2$. 
Hence we can now construct points $A, I, D$, with $I$ between $A$ and $D$ and $AI = d_1$, $ID = d_2$. We obtain points $K$ and $L$ as the intersections of of the circle with diameter $AI$ and the circle centered at $A$ with radius $e$. Thus we have constructed $\angle A$.
To complete the construction, we take $J$ to be the intersection of the circle with diameter $ID$ and the circle centered at $D$ of radius $f$. Then $DJ$ is the remaining side of the triangle.
Proof of the claims. It is well-known that $e = \frac{1}{2}(b + c - a)$, yielding the first claim. Likewise, the relation $BJ = \frac{1}{2}(a + c - b)$ leads to the second claim by writing $f = c' - BJ$. The third claim is obvious. The fourth follows by letting $r$ denote the inradius and writing $d_1^2 - e^2 = r^2 = d_2^2 - f^2$.
A: This construction is longer than in my other answer, but more geometrically motivated.
Let $d, x = b - b', y = c - c'$ be the given lengths.
Since $b'/c' = b/c$ and $b' + c' < b + c$, we have $x, y > 0$.
Without loss of generality, we may assume $x \geq y$. Since $x/y = b/c$ it follows that $b \geq c$.
A necessary condition (by the triangle inequality) is $x = b - b' < d$. Assume therefore that $x < d$. We will see conversely that this is sufficient and that in this case there is a unique triangle that works.
Analysis. Let $I$ be the incenter and $J, K, L$ the points of contact of the incircle with sides $BC$, $AC$, $AB$, respectively. Denote he inradius by $r$ and write $a = BC = b' + c'$, $e = AK = AL$, $f = DJ$. We have
$e = \frac{1}{2}(b + c - a) = \frac{1}{2}[(b - b') + (c - c')] = \frac{1}{2}(x + y),$
$f = c' - BJ = c' - \frac{1}{2}(a + c - b) = \frac{1}{2}[(b - b') - (c - c')] = \frac{1}{2}(x - y) \geq 0$
Let $\Gamma$ be the circle of radius $e$ centered at $A$, and $\Gamma'$ the (possibly degenerate) circle of radius $f$ centered at $D$. Then the radii $IJ = IK = IL = r$ of the incircle are respectively tangent to $\Gamma$ at $K$, to $\Gamma$ at $L$ and to $\Gamma'$ at $J$ (when $f > 0$). Thus $I$ has equal power $r^2$ with respect to $\Gamma$ and $\Gamma'$. Therefore $I$ is the intersection of $AD$ and the radical axis of $\Gamma$ and $\Gamma'$. 
Lastly, $J$, $K$ and $L$ are determined as points of contact of tangents to $\Gamma$ and $\Gamma'$ from $I$. The sides of $ABC$ are the lines $AK, AL, DJ$. (When $f = 0$, we have $J = D$ and the remaining side is the perpendicular to $AD$ through $D$. When $f > 0$, there are two choices of $J$, but they result in triangles congruent via reflection about $AD$.)
Construction.


*

*Let $AD$ be a segment of length $d$.

*Construct segments of length $e = \frac{1}{2}(x + y)$ and $f = e - y = \frac{1}{2}(x - y)$.

*Construct circle $\Gamma$ of radius $e$ with center $A$ and circle $\Gamma'$ of radius $f$ with center $B$. (Since $e + f = x < d$, the circles are outside each other.)

*Construct a point $M$ on the radical axis of $\Gamma$ and $\Gamma'$. Draw an arbitrary circle, with centre not on $AD$, which intersects each of $\Gamma$ and $\Gamma'$. Let $M$ be the intersection of the secants to $\Gamma$ and $\Gamma'$ determined by the four points of intersection.

*Construct the radical axis by either: (1) drawing the perpendicular to $AD$ through $M$, or (2) constructing a second point $M'$ of the radical axis as in point 4, and then drawing $MM'$.

*Construct the point $I$ of intersection of the radical axis with $AD$. (Since $I$ is on the radical axis, it has equal power with respect to each circle and it lies outside them.)

*Construct the points of intersection $K$ and $L$ of $\Gamma$ and the circle with diameter $AI$. (These exist because $I$ is outside $\Gamma$ and $A$ inside. Segments $IK = IL$ are tangent to $\Gamma$ at $K$ and $L$, respectively.) 

*Draw the circle $\Gamma''$ with center $I$ passing through $K$ and $L$. Let $J$ be its point of intersection with $\Gamma'$ that is on the same side of $AD$ as $L$. (Since $I$ is on the radical axis of $\Gamma$ and $\Gamma'$, the length of a tangent from $I$ to $\Gamma'$ is equal to $IK$. Hence $J$ exists, and $IJ$ is tangent to $\Gamma'$ at $J$ if $f > 0$. Circle $\Gamma''$ is tangent to $AK$ at $K$, to $AL$ at $L$ and, if $f > 0$, to $DJ$ at $J$.)

*Take the sides of triangle $ABC$ to be the lines $DJ$, $AK$ and $AL$. (If $f = 0$, then $J = D$; instead of $DJ$ take the tangent to $\Gamma''$ at $D$. In either case, $\Gamma''$ is tangent to all three lines. No two of the three lines can be parallel; in particular, $DJ$ and $AK$ cannot be parallel because the inequality $e > f$ precludes the possibility that $J$ and $K$ are diametrically opposite in $\Gamma''$.)
It follows from the construction that triangle $ABC$ has incircle $\Gamma''$, and that its points of contact with the sides are $J, K, L$. Line $AD$ passes through the incenter $I$, hence it is indeed a bisector. By construction, we have $AD = d$, $AK = AL = e = (x + y)/2$ and $DJ = f = (x - y)/2$. Hence by the previous analysis
$b - b' = AC - CD = e + f = x, \qquad c - c' = AB - BD = e - f = y,$
as required.
