直角三角形ABO,AO = H,BO = 2H,(H=20M)
抓到点为M,可知ABM为等腰三角形,过M作AB垂线,与AB 交点N为AB 中点.
AB = √5 H,BN=AB/2 = √5/2 H,
BNM与ABO相似有 BM / BN = AB / BO
BM = BN * AB / BO = 5/4H = 25 M.
MO = BO - BM = 15M
直角三角形ABO,AO = H,BO = 2H,(H=20M)
抓到点为M,可知ABM为等腰三角形,过M作AB垂线,与AB 交点N为AB 中点.
AB = √5 H,BN=AB/2 = √5/2 H,
BNM与ABO相似有 BM / BN = AB / BO
BM = BN * AB / BO = 5/4H = 25 M.
MO = BO - BM = 15M