Constructing a right triangle given $a+b-c$ and $\alpha$ The  exercise is to construct a right triangle given $a+b-c$ and $\alpha$.
I know we then have $\beta=90^\circ -\alpha$. I tried to draw the right triangle $\triangle ABC$ and find where I can use $a+b-c$. By adding $a$ to $b$ to get $\overline{DA}, |DA|=a+b$, I get an isosceles triangle $\triangle DCB$ so $|\angle CDB|=45^\circ$. I then subtract the length $c$ from $\overline{DA}$ and have $|DN|=a+b-c$. I thought I could construct the triangle $\triangle DNB$ so I could find $C$ as the intersection of the line $DN$ and the perpendicular bisector of $\overline{DB}$. However, I'm missing an element in order to be able to even construct $\triangle DNB$. Additionally, I tried adding lengths to the hypotenuse, but also without much success.

Alternative approach.
It is assumed that $c$ is the hypotenuse, that $a$ is the side opposite the angle $\alpha$, and that the value $(a+b-c)$ equals a given fixed value $T$.
Then, you have the following two equations in the two unknowns, $a,b$.

*

*$c^2 = (a + b - T)^2 = a^2 + b^2.$

*$\displaystyle \frac{a}{b} = \tan(\alpha)$.
Here, I am assuming that since $\alpha$ is known, that $\tan(\alpha)$ is known.

Once $a,b$ are solved, then $c$ is immediately solved by $a + b - c = T.$  Therefore, the problem reduces to using the above two equations to solve for $a,b$.
Using the first equation above, you have that
$$T^2 + 2ab = 2T(a+b). \tag1 $$
Let $R = \tan(\alpha).$ 
This implies that
$$a = bR \tag2 .$$
Using (1) and (2) above, you have that
$$T^2 + 2Rb^2 = 2Tb(R+1) \implies$$ 
$$b^2(2R) - b(2T[R+1]) + T^2 = 0. \tag3 $$
(3) above represents a quadratic in $b$.  Therefore,
$$b = \frac{1}{4R}\left[2T(R+1) \pm \sqrt{[2T(R+1)]^2 - 8RT^2}\right]. \tag4 $$
(4) above simplifies to
$$b = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right]. \tag5 $$
Temporarily, both roots in (5) will be carried, until $c$ is evaluated.  Since $c$ must be $> 0$, one of the roots will then be eliminated.
$$a = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right] \times R. \tag6 $$
Using that $c = a + b - T$, this implies that
$$c = \frac{T}{2R}\left[(R^2+1) \pm (R+1)\sqrt{R^2 + 1}\right]. \tag7 $$
At this point, since $R > 0 \implies \sqrt{R^2 + 1} < (R+1)$
you can infer that the smaller roots in (5), (6), and (7) are inappropriate.
You can then manually verify that the resulting values for $a,b,c$ satisfy $a^2 + b^2 = c^2.$

Edit
After reading the answer of Grab a Coffee, I was intrigued enough to sanity-check my answer.
$\displaystyle R = \frac{\sin(\alpha)}{\cos(\alpha)} \implies $

*

*$\displaystyle R^2 + 1 = \frac{1}{\cos^2(\alpha)}.$


*$\displaystyle (R + 1) = \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)}.$
Therefore,
$$c = \frac{T}{2} \times \frac{\cos(\alpha)}{\sin(\alpha)} \times 
\left[\frac{1}{\cos^2(\alpha)} +  \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)} \frac{1}{\cos(\alpha)}\right]. $$
This simplifies to
$$\frac{T}{2} \times \frac{1}{\sin(\alpha) \cos(\alpha)} \times [\sin(\alpha) + \cos(\alpha) + 1] 
\times \frac{\sin(\alpha) + \cos(\alpha) - 1}{\sin(\alpha) + \cos(\alpha) - 1}$$
$$ = \frac{T}{2} \times \frac{\sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha) - 1}{\sin(\alpha)\cos(\alpha) \times [\sin(\alpha) + \cos(\alpha) - 1]}$$
$$ = \frac{T}{2} \times \frac {2}{\sin(\alpha) + \cos(\alpha) - 1}.$$
