construct triangle given $b-c$, $r$ and $h_{b}$ As in title: the problem is to construct triangle given difference of sides $b$ and $c$, then in-circle radius $r$, and height $h_{b}$.
The problem is from a set of problems exercising various formulas for distances between points such as foots of in-circle and ex-circle etc.  One of such formulas claims that distance between foots of in-circle and ex-circle to side $a$ of given triangle (let's call these points $P$ and $P_{a}$) is equal to $b-c$, and that middle of side $a$ is also middle of segment between these two foots.  I'm pretty sure that this formula is to be used in given construction, so I'd start by drawing line segment $PP_{a}$ of size $b-c$, then constructing perpendicular to this segment in point $P$, and finding in-circle center $S$ at distance $r$ along this perpendicular from point $P$. Then I can draw in-circle, however the problem is now how to utilize height $h_{b}$...
 A: There may me more elegant solutions, but this works:
YO need to determine the length of side b.  This is determined in two steps:  First, find the lengths of the other sides in terms of b;  then set up an equation for b.
We are given b-c = Δ, so c is simply given as:
c = b – Δ     [1]
To determine a, note that the area of the triangle is given by one half the base times the height and also by one half the perimeter times the inradius.  To see how the latter is derived, consider a triangle in which the inscribed circle has been constructed.  Draw line segments form the center of this circle to the vertices of the triangle.  This divides the area of the original triangle into three components that are each one half of one side times the inradius.
Thus we have:
b*(h_b)=(a+b+c)r=(a+2b-Δ)r;
a= Δ + b((h_b)/r-2)    [2]
Note that (h_b)/r must be greater than 2 because of the triangle inequality.  Attempting to solve the system with (h_b)/r < 2 would still give a triangle, but it would be turned “inside out” with the “inscribed circle” lying outside the triangle; the circle would be tangent to side a and the extensions of sides b and c rather than all three sides proper.
Now equate the area as given by one half the base times the height with the area as given by Heron’s Formula:
2b*(h_b)=
[(a+b+c)(-a+b+c)(a-b+c)*(a+b-c)]^(1/2)   [3]
Substitute [1] and [2] into [3], square both sides, clear fractions and divide by a common factor of b2 that appears (b is of course nonzero).  This leads to a quadratic equation (which becomes linear in the specific case hb/r = 4:
(4r-(h_b))*((h_b)-2r)^2*b^2
+(12r-4(h_b))*((h_b)-2r)rΔ*b
-4*r^2*[((h_b)-2r)*Δ^2+(h_b)*r^2]=0      [4]
Equation [4] has a unique positive root when 2 < (h_b)/r < 4, and this will give a unique triangle meeting the given values of Δ, r and (h_b) .  For (h_b)/r > 4, we are not guaranteed a solution and with (h_b)/r > 4 there may be two roots.
A: Let $A'$ be the midpoint of $BC$. We will use the fact that, as you say, the tangent from $A'$ to the incircle has length $(b-c)/2$.
First draw a line $l$ that will be $AC$. (Don't place $A$ and $C$ yet.) Now select a point $I$ at a distance $r$ from $l$ to be the incentre. Draw the incircle, which is tangent to $l$ at a point $Q$.
The power of $A'$ with respect to the incircle must be $(b-c)^2/4$. The set of all points with this property is a circle concentric to the incircle. It can be constructed by selecting one point on $l$ at a distance $(b-c)/2$ from $Q$, and drawing the circle centred at $I$ through that point.
But $A'$ must also be located on a line parallel to $l$ at a distance $h_b/2$ from it, on the same side as the incircle. Construct $A'$ as either intersection of this line with the previously constructed circle.
Now line $BC$ can be constructed as a tangent to the incircle passing through $A'$ (two possibilities). Vertex $C$ is the intersection of this line with $l$, and $B$ is the reflection through $A'$ of $C$. Finally, $A$ can be obtained by drawing the other tangent to the incircle from $B$ and taking the intersection with $l$.
