14) em tham khảo ở đây nha
https://diendan.hocmai.vn/threads/c...-tiep-cua-cac-tam-giac-h.735843/#post-3696276
16 ) [tex]cosA=\frac{b^2+c^2-a^2}{2bc};sinA=\frac{a}{2R}[/tex]
[tex]=>cotA=\frac{b^2+c^2-a^2}{\frac{2abc}{4R}}=\frac{b^2+c^2-a^2}{2S}[/tex]
=> [tex]b^2+c^2-a^2=2(a^2+c^2-b^2+a^2+b^2-c^2)<=>b^2+c^2=5a^2[/tex]
gọi [TEX]G[/TEX] là giao điểm của 2 đường trung tuyến
vì BG vuông góc CG => [tex]BG^2+CG^2=BC^2<=>\frac{4}{9}\left ( \frac{a^2+c^2}{2}-\frac{b^2}{4}+\frac{a^2+b^2}{2}-\frac{c^2}{4} \right )=a^2<=>\frac{1}{9}(4a^2+c^2+b^2)=a^2<=>b^2+c^2=5a^2[/tex]
b) [tex]2(b^2+c^2-a^2)=(a^2+b^2-c^2+a^2+c^2-b^2)<=>b^2+c^2=2a^2[/tex]
[tex]\frac{m_{b}^2}{m_{c}^2}=\frac{c^2}{b^2}=\frac{2a^2+2c^2-b^2}{2a^2+2b^2-2c^2}<=>2c^2a^2-c^4=2b^2a^2-b^4<=>2a^2(b^2-c^2)-(b^2-c^2)(b^2+c^2)=0<=>b^2-c^2=0 \vee b^2+c^2=2a^2[/tex]
mà [tex]\frac{b^2}{c^2}\neq 1[/tex]=> [TEX]b^2+c^2=2a^2[/TEX]