∆ABC
AD=4x; DC=5x
EF hinh chieu D AB,AC
AE=AF=√(AD^2/2)=2√2.x
CF^2=DC^2 -DF^2=(25-8)x^2
CF=√17.x
AC=(2√2+√17) x
DF//BA=>CF/AC=CD/BC
BC=AC.CD/CF=[2√2+√17).5.x ]/√17
DE//AC=>BE/AB=DE/CA
(AB-BE)/AB=(AC-DE)/AC
AB=AC.AE/CF=[(2√2+√17).2√2.x ]/√17
Nguồn : Lazi