Ta có: AB2 + AC2 = BC2 (ĐL py-ta-go)
hay 172 + 122 = BC2
=> BC = \(\sqrt{17^2+12^2}\) = \(\sqrt{433}\) cm
Vì BM là đg phân giác của ^B
nên \(\frac{AM}{MC}=\frac{AB}{BC}\) hay \(\frac{AM}{MC}=\frac{17}{\sqrt{433}}\)
=> \(\frac{AM}{AM+MC}=\frac{17}{\sqrt{433}+17}\)
=> AM = \(\frac{17AC}{\sqrt{433}+17}\) = \(\frac{17.12}{\sqrt{433}+17}=\frac{204}{\sqrt{433}+17}\)
Có: AB2 + AM2 = BM2 (đ/l py-ta-go)
=> BM = \(\sqrt{AB^2+AM^2}\) = \(\sqrt{17^2+\left(\frac{204}{\sqrt{433}+17}\right)^2}\) = 17,84 cm