[imath]\left\{\begin{matrix}x-y-1=0\\2x+y-5=0\end{matrix}\right.\Rightarrow x=2;y=1[/imath]
[imath]\Rightarrow A(2,1)[/imath]
[imath]B(b,b-1); C(c,5-2c)[/imath]
[imath]\overrightarrow{MB}=(b-1,b); \overrightarrow{MC}=(c-1,6-2c)[/imath]
[imath]\overrightarrow{MB}=k\overrightarrow{MC}\Rightarrow (b-1)(6-2c)=b(c-1)[/imath]
[imath]\Rightarrow 6b-2bc-6+2c=bc-b\Rightarrow 7b+2c-6=3bc\Rightarrow b=\dfrac{6-2c}{7-3c}[/imath]
VTPT của [imath]d_1,d_2[/imath] lần lượt là [imath](1,-1); (2,1)[/imath]
[imath]\Rightarrow \cos \widehat{BAC}=\dfrac{\sqrt{10}}{10}[/imath]
[imath]BC^2=AB^2+AC^2-2AB.AC\cos \widehat{BAC}[/imath]
[imath]9AB^2=AB^2+AC^2-AB.AC\dfrac{\sqrt{10}}{5}\Rightarrow AB=\dfrac{\sqrt{10}}{8}AC[/imath]
[imath]\overrightarrow{AB}=(b-2,b-2); \overrightarrow{AC}=(c-2,4-2c)[/imath]
Suy ra [imath]2|b-2|=\dfrac{\sqrt{10}}{8}\sqrt{(c-2)^2+(4-2c)^2}[/imath]
[imath]\Rightarrow 64(b-2)^2=5(5c^2-20c+20)[/imath]
[imath]\Rightarrow 64\left(\dfrac{4c-8}{3c-7}\right)^2=5(5c^2-20c+20)[/imath]
[imath]\Rightarrow c=2 \:(l); c=\dfrac{1}5\: (n); c=\dfrac{67}{15}\: (n)[/imath]
TH1: [imath]C(\dfrac{1}5,\dfrac{23}5); B(\dfrac{7}8; \dfrac{-1}8)[/imath]
TH2: [imath]C(\dfrac{67}{15}; \dfrac{-59}{15}); B(\dfrac{11}{24}; \dfrac{-13}{24})[/imath]
Em tự viết pt đt nha