<s>[INST] Find $\angle B$ if $AD=\frac{abc}{b^2-c^2}$ If AD is median and $AD=\frac{abc}{b^2-c^2}$ $[b>c]$ and $\angle C=23^{\circ} $.
Find $\angle B$
Is this information sufficient to find $\angle B$? I tried using sine rule in triangle $ADC$ and Appolonius theorem in triangle $ABC$ but can't find way to reach value of $\angle B$.
Can someone suggest something?
 [/INST] By the theorem of cosines, $c^2=a^2+b^2-2ab\cos\gamma$. Then $AD=\frac{bc}{2b\cos\gamma-a}$. It follows that $t=\frac{a}b < 2\cos\gamma$.
Further, the median length of the formula, $AD^2=(b^2+c^2)/2-a^2/4=a^2/4+b^2-ab\cos\gamma$. This expression equate to $\frac{b^2c^2}{(2b\cos\gamma-a)^2}=\frac{b^2(a^2+b^2-2ab\cos\gamma)}{(2b\cos\gamma-a)^2}$. Divide both expressions on $b^2$. Coming to the equation $t^2/4-t\cos\gamma+1=\frac{t^2-2t\cos\gamma+1}{(t-2\cos\gamma)^2}$.
As a result, there is an equation of the 4th degree, which has two roots. It was solved numerically, but in the process of writing became clear unexpected thing:it turns out, it lay on quadratic factors for any value of the angle $\gamma$. Namely, $(t^2-6t\cos\gamma+8\cos^2\gamma-2)(t^2-2t\cos\gamma+2)=0$. It is clear that the second factor has a negative discriminant and solutions does not, and for the first turns $t=3\cos\gamma\pm\sqrt{2+\cos^2\gamma}$. The second value does not satisfy the inequality and the first approach.
It turns out $t=1.074111471$ (approximately). The sides are proportional to the numbers $a=1.074111468$, $b=1$, $c=0.419840216$, and the angles are equal $\beta\approx68.53891392$ (degrees), and $\alpha\approx88.46108608$. 
Acute-angled triangle is obtained; the solution is unique.
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