We have right angle $ABC$ where $AB$ is hypotenuse, and angle bisector $CD$ we know that $AB=c$, $CD=u$. Express height depending on the $c, u$, and what is the condition between $c,u$ that the triangle $ABC$ exist?
I have some problems here I tried to enter new viariable $a,b$ and use Pythagorean theorem and the bisector theorem, but without success.
Let $BC=x$. Noting that $AC=\sqrt{c^2-x^2}$, let us consider the areas. Let $(A)$ be the area of a figure $A$.
$$(\text{a triangle}\ ABC)=(\text{a triangle}\ BCD)+(\text{a triangle}\ ACD)$$ $$\Rightarrow\ \frac 12x\sqrt{c^2-x^2}=\frac 12xu\sin\frac{\pi}{4}+\frac 12u\sqrt{c^2-x^2}\sin\frac{\pi}{4}$$ $$\Rightarrow\ 2x\sqrt{c^2-x^2}=\sqrt2 xu+\sqrt 2u\sqrt{c^2-x^2}$$ $$\Rightarrow\ \sqrt{c^2-x^2}(2x-\sqrt2u)=\sqrt 2xu$$ $$\Rightarrow\ (c^2-x^2)(2x-\sqrt 2u)^2=2x^2u^2$$ $$\Rightarrow\ 2x^4-2\sqrt2ux^3+(u^2-2c^2)x^2+2\sqrt 2uc^2x-c^2u^2=0$$ $$\Rightarrow\ (x-c)(x+c)(\sqrt2x-u)^2=0$$ $$\Rightarrow\ x=c\ \text{or}\ x=-c\ \text{or}\ x=\frac{u}{\sqrt2}.$$
By the way, $x=\pm c$ are not what we want because of $AC=\sqrt{c^2-x^2}.$ Hence, the answer should be $x=\frac{u}{\sqrt2}$.
In addition, the condition that a triangle can exist is that $$c^2-x^2\gt0\ \iff\ c^2-\frac{u^2}{2}\gt0\ \iff\ u^2\lt 2c^2\ \iff\ u\lt \sqrt 2c\ \iff\ \frac{u}{c}\lt \sqrt 2.$$
Considering ares is sometimes useful.