Construction of a right triangle

Let $ABC$ be the triangle you want to construct, with $\angle A =90^\circ$.

Then you are given $AB+AC$ and $BC$.

Extend $BA$ past $A$ to $BB'$ by a length equal to $AC$, that is $AB'=AC$. Then, the triangle $ACB'$ is a right isosceles triangle.

This means that in the triangle $BB'C$ you know $BB'=AB+AC$, $BC$ and the angle $B'=45^\circ$.

This suggests how you can construct it: construct the triangle $BB'C$, and then construct the height from $C$. The leg of the height will be $A$.

Since you are constructing $BB'C$ by $SSA$, there should be two solutions for $C$.

Here is the actual construction:

Start by drawing an angle of $45^\circ$. Denote the vertex of the angle $B'$.

On one side pick a point $B$ so that $BB'=AC+AB$.

Next draw a circle of centre $B$ and radius $BC$. This will intersect the other ray of the angle in two points $C_1, C_2$. Pick the one which makes the angle $CBB'$ acute.

P.S. I always solve the problems the way i.m.s. did, I like more the algebraic approach, but since you mention that this is high school level you are probably looking for the geometric approach.

You know the hypotenuse length $c$, and the sum of the lengths of the legs $u=a+b$. Thus you also know $v=u^2-c^2=a^2+b^2+2ab-a^2-b^2=2ab$. Then $b=u-a=u-v/2b$ so b satisfies the quadratic equation $b^2-ub+v/2=0$ which can be constructed using ruler and compass.