A right triangle $ABC$ is given with right angle at $C$. If $AB=a$ and the angle bisector of $A$ is $AL=l$, find the area of the triangle $ABC$.
The angles of the triangle $ABL$ are $\dfrac{\alpha}{2}, 90-\alpha$ and $90+\dfrac{\alpha}{2}$, where $\alpha$ is the angle $BAC$.
The law of sines gives $$\dfrac{\sin\left(90+\dfrac{\alpha}{2}\right)}{\sin(90-\alpha)}=\dfrac{a}{l}$$ which is equivalent to $$\dfrac{\cos\dfrac{\alpha}{2}}{\cos\alpha}=\dfrac{a}{l}$$ If we use that $\cos\alpha=2\cos^2\dfrac{\alpha}{2}-1$, the last equality becomes an equation for $\cos\dfrac{\alpha}{2}=x>0$. We would have $$\cos\dfrac{\alpha}{2}=\dfrac{a}{l}\left(2\cos^2\dfrac{\alpha}{2}-1\right) \\ \dfrac{2a}{l} x^2-x-\dfrac{a}{l}=0 \\ 2ax^2-lx-a=0$$ The positive solution is $x=\cos\dfrac{\alpha}{2}=\dfrac{l+\sqrt{l^2+8a^2}}{4a}$.
By definition $\cos\dfrac{\alpha}{2}$ is $\dfrac{AC}{AL}$, so for $AC=b$ I got $$b=\dfrac{l^2+l\sqrt{l^2+8a^2}}{4a}$$ From here I don't see anything else except the Pythagorean theorem for $BC$ but it's pretty messy. I wish I could see something better. I tried letting Wolfram Alpha do the simplification, which probably avoids errors, but it's still ugly:
(l (√(8 a^2 + l^2) + l) √(a^2 - (l^2 (√(8 a^2 + l^2) + l)^2)/(16 a^2)))/(8 a)
I have tried to find a better way, but I keep coming back to expressions involving $b$ and the Pythagorean Theorem. For instance, the area of $\triangle ABC$ can be expressed as $\dfrac{c(a+b)}{2}$ where $c = |CL|$; but this doesn't simplify any more neatly.
HINT.-Be the right triangle $\triangle {ABC}$ as in the problem and let $a,b$ the sides forming the vertex $A$ where is the angle$2\alpha$. We have $$b=l\cos(\alpha)=a\cos(2\alpha)\iff a(2\cos^2(\alpha)-1)=l\cos(\alpha)$$ $$2a\cos^2(\alpha)-l\cos(\alpha)-a=0$$ and $$\cos(\alpha)=\frac{l\pm\sqrt{l^2+8a^2}}{4a}$$ Finally the area $S$ is equal to $$\dfrac12ab\sin(2\alpha)=ab\cos(\alpha)\sqrt{1-\cos^2(\alpha)}$$