BC = \(\sqrt{8^2+6^2}\)= 10 cm
trung truyến AM = BC/2 = 5cm
AG = 2AM/3 = 10/3 cm.
trung tuyến BN = \(\sqrt{\frac{2BC^2+2BA^2-AC^2}{4}}\)= \(\sqrt{\frac{2\left(10^2+6^2\right)-8^2}{4}}\)
BG = 2BN/3
trung tuyến CP = \(\sqrt{\frac{2BC^2+2AC^2-AB^2}{4}}\)= \(\sqrt{\frac{2\left(10^2+8^2\right)-6^2}{4}}\)
BG = 2CP/3