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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.

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  • $\begingroup$ What is $CD$ bisecting? $AB$ or $\angle C?$ Which altitude are you trying to express-the one from $C$? $\endgroup$ – Ross Millikan Dec 17 '13 at 20:39
  • $\begingroup$ $CD$ devide $\angle C$ on equal parts $\endgroup$ – Mark Dec 17 '13 at 20:41
  • $\begingroup$ If C is right angle, u=c/2, so your question may be not correct. $\endgroup$ – chenbai Dec 18 '13 at 3:12
  • $\begingroup$ no, $u\neq\frac{c}{2}$ u is angle bisector section $\endgroup$ – Mark Dec 18 '13 at 12:50
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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.

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  • $\begingroup$ but, why are you assume that $CD$ is median? $\endgroup$ – Mark Dec 18 '13 at 17:25
  • $\begingroup$ I misunderstood it. Now I hope I'm not mistaken. $\endgroup$ – mathlove Dec 19 '13 at 4:58

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