Question

菱形abcd中,对角线ac=8 bd=6,dh垂直ab于点h,且交ac于点g,求hg长度

Answer 1

菱形abcd边长=√[(ac/2)^2+(bd/2)^2] =√[(8/2)^2+(6/2)^2] = 5

增加对角线AC、BD的标记点为O

△BDH ∽△ABO

OA/DH=AB/BD

DH = OA*BD / AB = 8/2 * 6 / 5 = 24/5

△DGO ∽△ABO

DG/AB = OD/OA

DG = AB*OD / OA = 5*6/2 / 8/2 = 15/4

hg长度＝DH - DG = 24/5-15/4 =21/20