已知 ΔABC 中从 A 点出发的中线 ma=11 ，角平分线 ba=7 ，高 ha=5 ，求 ΔABC 的面积

王守恩

接 6#。——还是啰嗦,  不知标准答案是怎样的？

\(5^2 + z^2 = k^2y^2, (1)\)

\(5^2 + (y + z)^2 = 7^2, (2)\)

\(5^2 + (x + z)^2 = 11^2, (3)\)

\(5^2 + (2 x + z)^2 = k^2(2 x - y)^2,(4)\)

\(由(2)—(y+z)^2=24\)

\(由(3)—(x+z)^2=((x-y)+(y+z))^2=4*24,可得x-y=y+z=24.可得x-2y=z\)

\(k^2=\frac{5^2+z^2}{y^2}=\frac{5^2+(2x+z)^2}{(2x-y)^2}\)

\(k^2=\frac{5^2+(x-2y)^2}{y^2}=\frac{5^2+(3x-2y)^2}{(2x-y)^2}\)

\(k^2=\frac{5^2+(2(x-y)-x)^2}{y^2}=\frac{5^2+(2(x-y)+x)^2}{((x-y)+x)^2}\)

\(k^2=\frac{5^2+(2\sqrt{24}-x)^2}{(\sqrt{24}-x)^2}=\frac{5^2+(2\sqrt{24}+x)^2}{(\sqrt{24}+x)^2}\)

\(即:\frac{(\sqrt{24}-x)^2}{(\sqrt{24}+x)^2}=\frac{25+2\sqrt{24}-x)^2}{25+2\sqrt{24}+x)^2}利用合分比\frac{u}{v}=\frac{p}{q}可以有\frac{u+v}{u-v}=\frac{p+q}{p-q}\)

\(\frac{(\sqrt{24}-x)^2+(\sqrt{24}+x)^2}{(\sqrt{24}-x)^2-(\sqrt{24}+x)^2}=\frac{25+(2\sqrt{24}-x)^2+25+(2\sqrt{24}+x)^2}{25+(2\sqrt{24}-x)^2-25-(2\sqrt{24}+x)^2}\)

\(\frac{2(24+x^2)}{-4\sqrt{24}x}=\frac{2*25+2(4*24+x^2)}{-4*2\sqrt{24}x}\)

\(\frac{24+x^2}{\sqrt{24}}=\frac{25+4*24+x^2}{2\sqrt{24}}\)

\(48+2x^2=25+4*24+x^2\)

\(x^2=73\)