Если CD -биссектриса, то ∠BCD=∠DCA=∠C/2=90°/2=45°
∠CDA=180°-(∠ACD+∠DAC)=180°-(45°+15°)=120°
∠BDC=180°-∠CDA=180°-120°=60°
Найдем х по теореме синусов:
![\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%7D%7BsinA%7D+%3D+%5Cfrac%7Bb%7D%7BsinB%7D+%3D+%5Cfrac%7Bc%7D%7BsinC%7D+)
в треугольнике АСD:
![\frac{AC}{sin120} = \frac{x}{sin45} \\ \\ \frac{ \sqrt{3} }{ \frac{ \sqrt{3} }{2} }= \frac{x}{ \frac{ \sqrt{2} }{2} } \\ \\ 1= \frac{x}{ \sqrt{2} } \\ \\ x= \sqrt{2}](https://tex.z-dn.net/?f=+%5Cfrac%7BAC%7D%7Bsin120%7D+%3D+%5Cfrac%7Bx%7D%7Bsin45%7D++%5C%5C++%5C%5C++%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B++%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D++%7D%3D+%5Cfrac%7Bx%7D%7B+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+%7D+++%5C%5C++%5C%5C+1%3D+%5Cfrac%7Bx%7D%7B+%5Csqrt%7B2%7D+%7D++%5C%5C++%5C%5C+x%3D+%5Csqrt%7B2%7D+)
Пусть BD=a, тогда теорема синуса для ΔBCD:
![\frac{a}{sin45}= \frac{y}{sin60}](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%7D%7Bsin45%7D%3D+%5Cfrac%7By%7D%7Bsin60%7D++)
с другой стороны ΔАВС - прямоугольный, значит для него выполняется теорема Пифагора:
![AB^2=BC^2+AC^2 \\ \\ (a+x )^2=y^2+( \sqrt{3} )^2 \\ \\](https://tex.z-dn.net/?f=+AB%5E2%3DBC%5E2%2BAC%5E2+%5C%5C++%5C%5C+%28a%2Bx+%29%5E2%3Dy%5E2%2B%28+%5Csqrt%7B3%7D+%29%5E2+%5C%5C++%5C%5C)
учитывая, что х=√2
![(a+ \sqrt{2} )^2=y^2+3](https://tex.z-dn.net/?f=%28a%2B+%5Csqrt%7B2%7D+%29%5E2%3Dy%5E2%2B3)
получается система из 2 уравнений:
![\left \{ {{ \frac{a}{sin45}= \frac{y}{sin60}} \atop {(a+ \sqrt{2} )^2=y^2+3}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7Ba%7D%7Bsin45%7D%3D+%5Cfrac%7By%7D%7Bsin60%7D%7D+%5Catop+%7B%28a%2B+%5Csqrt%7B2%7D+%29%5E2%3Dy%5E2%2B3%7D%7D+%5Cright.+)
из первого выражаем а и подставляем во второе:
![\frac{a}{sin45}= \frac{y}{sin60} \\ \\ \frac{a}{ \sqrt{2} /2}= \frac{y}{ \sqrt{3}/2 } \\ \\ \frac{a}{ \sqrt{2} }= \frac{y}{ \sqrt{3} } \\ \\ a= \frac{y \sqrt{2} }{ \sqrt{3} } \\ \\](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%7D%7Bsin45%7D%3D+%5Cfrac%7By%7D%7Bsin60%7D+%5C%5C++%5C%5C++%5Cfrac%7Ba%7D%7B+%5Csqrt%7B2%7D+%2F2%7D%3D+%5Cfrac%7By%7D%7B+%5Csqrt%7B3%7D%2F2+%7D+%5C%5C++%5C%5C+%5Cfrac%7Ba%7D%7B+%5Csqrt%7B2%7D+%7D%3D+%5Cfrac%7By%7D%7B+%5Csqrt%7B3%7D+%7D++%5C%5C++%5C%5C+a%3D+%5Cfrac%7By+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B3%7D+%7D++%5C%5C++%5C%5C+)
![(a+ \sqrt{2} )^2=y^2+3 \\ \\ a^2+2 \sqrt{2}a +2=y^2+3 \ \ =\ \textgreater \ \ a= \frac{y \sqrt{2} }{ \sqrt{3} } \\ \\ \frac{2y^2}{3}+ 2 \sqrt{2}* \frac{y \sqrt{2} }{ \sqrt{3} }+2=y^2+3 \ \ |*3 \\ \\ 2y^2+4 \sqrt{3} y+6=3y^2+9 \\ \\ 3y^2-2y^2-4 \sqrt{3} y-6+9=0 \\ \\ y^2-4 \sqrt{3} +3=0 \\ \\ D=16*3-4*3=36=6^2 \\ \\ y= \frac{4 \sqrt{3}+6 }{2} =2 \sqrt{3} +3 \\ \\ OTBET: \ x= \sqrt{2} ; \ \ y=2 \sqrt{3} +3](https://tex.z-dn.net/?f=%28a%2B+%5Csqrt%7B2%7D+%29%5E2%3Dy%5E2%2B3++%5C%5C++%5C%5C+a%5E2%2B2+%5Csqrt%7B2%7Da+%2B2%3Dy%5E2%2B3+%5C+%5C+%3D%5C+%5Ctextgreater+%5C++%5C+a%3D+%5Cfrac%7By+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B3%7D+%7D++%5C%5C++%5C%5C++%5Cfrac%7B2y%5E2%7D%7B3%7D%2B+2+%5Csqrt%7B2%7D%2A++%5Cfrac%7By+%5Csqrt%7B2%7D+%7D%7B+%5Csqrt%7B3%7D+%7D%2B2%3Dy%5E2%2B3+%5C+%5C+%7C%2A3++%5C%5C++%5C%5C+2y%5E2%2B4+%5Csqrt%7B3%7D+y%2B6%3D3y%5E2%2B9+%5C%5C+%5C%5C+3y%5E2-2y%5E2-4+%5Csqrt%7B3%7D+y-6%2B9%3D0+%5C%5C++%5C%5C+y%5E2-4+%5Csqrt%7B3%7D+%2B3%3D0+%5C%5C++%5C%5C+D%3D16%2A3-4%2A3%3D36%3D6%5E2+%5C%5C++%5C%5C+y%3D+%5Cfrac%7B4+%5Csqrt%7B3%7D%2B6+%7D%7B2%7D+%3D2+%5Csqrt%7B3%7D+%2B3+%5C%5C++%5C%5C+OTBET%3A+%5C+x%3D+%5Csqrt%7B2%7D+%3B+%5C+%5C+y%3D2+%5Csqrt%7B3%7D+%2B3+)