三角形abc中,ab=6,ac=4,bc=8,bd=dc,ae垂直bc,求ad;ae,dede

Question

三角形abc中,ab=6,ac=4,bc=8,bd=dc,ae垂直bc,求ad;ae,de de

Answer 1

设AE=x
则BE=√(6^2-x^2),EC=√(4^2-x^2)
BE+EC=BC
√(6^2-x^2) + √(4^2-x^2) = 8
√(6^2-x^2) = 8 - √(4^2-x^2)
6^2-x^2 = 8^2 -16√(4^2-x^2) + 4^2-x^2
16√(4^2-x^2) = 8^2+4^2-6^2
√(4^2-x^2) = 11/4
x^2 = 4^2-121/4^2 = 135/16
x = 3√15/4
即：AE = 3√15/4
BE=√(6^2-x^2) = √(6^2-135/16) = 21/4
BD=DC=BC/2=8/2=4
DE=BE-BD=21/4-4=5/4
AD = √(AE^2+DE^2) = √(135/16+25/16) = √10